NCERT Solutions for Class 11 Maths Chapter 4 - Mathematical Induction

Exercise 4.1

Prove the following by using the principle of mathematical induction for all n ∈ N:

$$\textbf{1. 1 + 3 + 3}^\textbf{2}\textbf{ + … + 3}^\textbf{n – 1}=\frac{\textbf{3}^{\textbf{n}}\textbf{-1}}{\textbf{2}}.$$

$$\textbf{Sol.}\space\text{Let P(n) : 1 + 3 + 3}^{2}+ ... +3^{n-1}=\frac{3^{n}-1}{2}\\\text{ for n = 1, } $$

we get

$$\text{P(1)}:\frac{3^1-1}{2}=\frac{3-1}{2}=\frac{2}{2}=1;$$

P(1) statement is true

For consider that P(n) statement is true for n = k,

$$\text{P(k) : 1 + 3 + 3}^{2} + ... +3^{k-1}=\frac{3^{k}-1}{2}\space\text{...(i)}$$

For now, let us prove that P(n) will also true for
n = k + 1,

P(k + 1) : 1 + 3 + 3² + … + 3^{k – 1} + 3^{k + 1 – 1}

= 1 + 3 + 3² + … + 3^{k – 1} + 3^k

$$=\frac{3^{k}-1}{2} + 3^{k}\space[\text{by (i)}]$$

Taking L.C.M.

$$=\frac{3^{k} + 2×3^{k}-1}{2}\\=\frac{3(3^{k})-1}{2}\\=\frac{3^{k+1}-1}{2}$$

∴ P(k + 1) is true, when P(k) is true.

Hence, from the principle of mathematical induction, the statement P(n) is true ∀ n ∈ N.

$$\textbf{2. 1}^{\textbf{3}} \textbf{+ 2}^\textbf{3} + \textbf{3}^{\textbf{3}}\textbf{ + … + n}^\textbf{3} =\bigg[\frac{\textbf{n(n+1)}}{\textbf{2}}\bigg]^{\textbf{2}}\\\textbf{Sol.}\space\text{Let P(n) : 1}^3 + 2^3 + 3^3 + … + n^3=\\\bigg[\frac{n(n+1)}{1}\bigg]^{2}$$

for n = 1, we get

$$\text{P(1)}:\bigg[\frac{1(1+1)}{2}\bigg]^{2}=\bigg[\frac{1×2}{2}\bigg]^{2}$$

= 1² = 1 = 1³;

P(1) statement is true for n = 1.

For consider that P(n) statement is true for n = k,

$$\text{P(k)}:1^{3} + 2^{3} + 3^{3} +...+ k^{3}\\=\bigg[\frac{k(k+1)}{2}\bigg]^{2}\space...(i)$$

For now, we’ll prove that P(n) statement is also
true for n = k + 1

P(k + 1) : 1³ + 2³ + … + k³ + (k + 1)³

$$=\bigg[\frac{k(k+1)}{2}\bigg]^{2} + (k+1)^{3}$$

[by equation (i)]

$$=\frac{k^{2}(k+1)^{2}}{4}+(k+1)^{3}\\=\frac{k^{2}(k+1)^{2} + 4(k+1)^{3}}{4}\\=\frac{(k+1)^{2}(k^{2} + 4k +4)}{4}\\=\frac{(k+1)^{2}(k+2)^{2}}{4}$$

{∵ (a + b)² = a² + b² + 2ab}

$$=\frac{(k+1)^{2}[(k+1)+1]^{2}}{4}\\=\bigg[\frac{(k+1)[(k+1)+1]}{2}\bigg]^{2}$$

∴ P(k + 1) is true, when P(k) is true.

Hence, from the principle of mathematical induction, the statement P(n) is true ∀ n ∈ N.

$$\textbf{3. 1+}\frac{\textbf{1}}{\textbf{(1+2)}}+ \frac{\textbf{1}}{\textbf{(1+2+3)}}\textbf{+...}\\\textbf{+}\frac{\textbf{1}}{\textbf{(1+2+...+n)}}\textbf{=}\frac{\textbf{2n}}{\textbf{(n+1)}}.\\\textbf{Sol.}\space\text{Let P(n) : 1}+\frac{1}{1+2}+...\\+\frac{1}{1+2+...+n}=\frac{2n}{n+1}$$

for n = 1, we get

$$\text{P(1)}=\frac{2}{1+1}=\frac{2}{2}=1;$$

P(1) statement is true.

For consider that P(n) statement is true for n = k

$$\text{P(k)}:1+\frac{1}{1+2}+...+\frac{1}{1+2+...+K}\\=\frac{2k}{k+1}$$

$$\Rarr\space1+\frac{1}{1+2}+...+\frac{1}{\frac{k(k+1)}{2}}\\=\frac{2k}{k+1}\\\bigg(\because\space 1+2+...+n=\frac{n(n+1)}{2}\bigg)\\\Rarr\space1+\frac{1}{1+2}+...+\frac{2}{k(k+1)}\\=\frac{2k}{k+1}\space\text{...(i)}$$

For now, we ’ll show that P(n) is also true for n = k + 1,

$$\text{P(k+1)}: 1+\frac{1}{1+2}+...+\frac{2}{k(k+1)}\\+\frac{2}{(k+1)(k+2)}\\=\frac{2k}{k+1}+\frac{2}{(k+1)(k+2)}\\\text{[\text{by (i)}]}\\=\frac{2k(k+2)+2}{(k+1)(k+2)}\\=\frac{(2k^{2}+4k)+2}{(k+1)(k+2)}\\=\frac{2(k^{2}+2k+1)}{(k+1)(k+2)}$$

Further simplification,

$$=\frac{2(k+1)^{2}}{(k+1)(k+2)}$$

{∵ (a + b)² = a² + b² + 2ab}

$$=\frac{2(k+1)}{(k+2)}$$

At last, we get

$$=\frac{2(k+1)}{(k+1)+1}$$

∴ P(k + 1) is true; when P(k) is true.

Hence, from the principle of mathematical induction, the statement P(n) is true ∀ ∈ N.

4. 1.2.3 + 2.3.4 + … + n(n + 1) (n + 2)

$$=\frac{\textbf{n(n+1)(n+2)(n+3)}}{\textbf{4}}.$$

Sol. Let P(n) : 1.2.3 + … + n(n + 1) (n + 2)=

$$=\frac{n(n+1)(n+2)(n+3)}{4}.$$

for n = 1, we get

$$\text{P(1)}:\space=\frac{1(1+1)(1+2)(1+3)}{4}\\=\frac{1×2×3×4}{4}$$

= 1.2.3; P(1) statement is true.

For consider that P(n) statement is true for n = k,

P(k) : 1.2.3 + 2.3.4 + … + k(k + 1) (k + 2)

$$=\frac{k(k+1)(k+2)(k+3)}{4}\space\text{...(i)}$$

For now, let us prove that P(n) is also true for n = k + 1,

P(k + 1) : 1.2.3 + 2.3.4 + … + k(k + 1) (k + 2) + (k + 1) (k + 2) (k + 3)

$$=\frac{k(k+1)(k+2)(k+3)}{4}+\text{(k+1)(k+2)(k+3)}$$

[by equation (i)]

Taking L.C.M.

$$=\\\frac{k(k+1)(k+2)(k+3)+4(k+1)(k+2)(k+3)}{4}$$

Further calculation

$$=\frac{(k+1)(k+2)(k+3)(k+4)}{4}\\=\frac{(k+1)[(k+1)+1][(k+1)+2][(k+1)+3]}{4}$$

∴ P(k + 1) is true, when P(k) is true.

Hence, from the principle of mathmatical induction, the statement P(n) is true, ∀ n ∈ N

5. 1.3 + 2.3² + 3.3³ + … + n.3ⁿ

$$\textbf{=}\frac{\textbf{(2n-1)3}^\textbf{{n+1}}\textbf{+3}}{\textbf{4}}.$$

Sol. Let P(n) : 1.3 + 2.3² + … + n.3ⁿ =

$$\frac{(2n-1)3^{n+1}+3}{4}$$

for n = 1, we get

$$\text{P(1):}\space\frac{[2(1)-1]3^{1+1}+3}{4}\\=\frac{(2-1)3^{2}+3}{4}\\=\frac{3^{2}+3}{4}=\frac{9+3}{4}\\=\frac{12}{4}=3=1.3;$$

P(1) statement is true.

For consider that P(n) statement is true for n = k,

P(k) : 1.3 + 2.3² + … + k.3^k

$$=\frac{(2k-1)3^{k+1}+3}{4}\space\text{...(i)}$$

For now, we’ll prove that P(n) is also true for
n = k + 1,

P(k + 1) : 1.3 + 2.3² + … + k.3^k + (k + 1).3^{k + 1}

$$=\frac{(2k-1)3^{k+1}+3}{4}+(k+1)3^{k+1}\\\space\text{[by (i)]}$$

Taking L.C.M.

$$=\frac{(2k-1)3^{k+1}+3+4(k+1)3^{k+1}}{4}$$

Further calculation

$$=\frac{3^{k+1}(2k-1+4k+4)+3}{4}\\=\frac{3^{k+1}(6k+3)+3}{4}\\=\frac{3^{k+1}.3(2k+1)+3}{4}\\=\frac{3^{(k+1) +1}.(2k+2-1)+3}{4}$$

Atlast, we get

$$=\frac{3^{(k+1)+1}[2(k+1)-1]+3}{4}$$

∴ P(k + 1) is true when P(k) is true.

Hence, from the principle of mathematical induction, the statement P(n) is true ∀ n ∈ N.

6. 1.2 + 2.3 + 3.4 + … + n(n + 1) =

$$\bigg[\frac{\textbf{n(n+1)(n+2)}}{\textbf{3}}\bigg].$$

Sol. Let P(n) : 1.2 + 2.3 + … + n(n + 1) =

$$\frac{n(n+1)(n+2)}{3}$$

for n = 1, we get

$$\text{P(1)}:\frac{1(1+1)(1+2)}{3}\\=\frac{1×2×3}{3}=1.2;$$

P(1) statement is true.

For consider that P(n) statement is true for n = k

P(k) : 1.2 + 2.3 + … + k(k + 1) =

$$\frac{k(k+1)(k+2)}{3}\text{...(i)}$$

For now, let us show P(n) will also be true for
n = k + 1,

P(k + 1) : 1.2 + 2.3 + … + k(k + 1) + (k + 1) (k + 2)

$$=\frac{k(k+1)(k+2)}{3}+(k+1)(k+2)$$

[by (i)]

Taking L.C.M

$$=\frac{k(k+1)(k+2)+3(k+1)(k+2)}{3}$$

Further calculation, we get

$$=\frac{(k+1)(k+2)(k+3)}{3}\\=\frac{(k+1)[(k+1)+1][(k+1)+2]}{3}$$

∴ P(k + 1) is true when P(k) is true.

Hence, from the mathematical induction, the statement P(n) is true ∀ n ∈ N.

7. 1.3 + 3.5 + 5.7 + … + (2n – 1) (2n + 1)

$$=\frac{\textbf{n}(\textbf{4n}^{\textbf{2}}+\textbf{6n - 1})}{\textbf{3}}.$$

for n = 1, we get

$$\text{P(1):}\frac{1[4(1)^{2} + 6(1)-1]}{3}\\=\frac{4.1+6-1}{3}=\frac{9}{3}=3$$

= 1.3; P(1) statement is true.

For now, consider P(n) will be true for n = k,

P(k) : 1.3 + 3.5 + … + (2k – 1) (2k + 1)

$$=\frac{k(4k^{2}+6k-1)}{3}\space\text{...(i)}$$

Now, let us prove that P(n) will also true for n = k + 1.

P(k + 1) : 1.3 + 3.5 + … + (2k – 1) (2k + 1) + [2(k + 1) – 1)] [2(k + 1) + 1]

$$=\frac{k(4k^{2}+6k-1)}{3}\\+[2k+2-1][2k+2+1]$$

[by using equation (i)]

$$=\frac{k(4k^{2}+6k-1)}{3}\\+(2k+1)+(2k+3)$$

On further calculation

$$=\frac{k(4k^{2}+6k-1)+3(4k^{2}+6k+2k+3)}{3}\\=\frac{4k^{3}+6k^{2}-k+12k^{2}+24k+9}{3}\\=\frac{4k^{3}+18k^{2}+23k+9}{3}$$

by using remainder theorem

$$=\frac{(k+1)(4k^{2}+14k+9)}{3}\\=\frac{(k+1)[4k^{2}+8k+4+6k+5]}{3}\\=\frac{(k+1)[4(k^{2}+2k+1)+6k+6-1]}{3}\\=\frac{(k+1)[4(k+1)^{2}+6(k+1)-1]}{3}$$

∴ P(k + 1) is true, when P(k) is true.

Hence, from the principle of mathematical induction, the statement P(n) is true ∀ n ∈ N.

8. 1.2 + 2.2² + 3.2³ + … + n.2ⁿ = (n – 1) 2^{n + 1} + 2.

Sol. Let P(n) : 1.2 + 2.2² + 3.2³ + … + n.2ⁿ= (n – 1)2^{n + 1} + 2

for n = 1, we get

P(1) : (1 – 1)2^{1 + 1} + 2 = 0.2²+ 2 = 2;

P(1) is statement is true.

For so, let us prove P(n) statement is true for
n = k,

P(k) : 1.2 + 2.2² + 3.2³ + … + k.2^k = (k – 1)2^{k + 1} + 2 …(i)

For n = k + 1, considering P(n) is also true.

P(k + 1) : 1.2 + 2.2² + 3.2³ + … + k.2^k + (k + 1)2^{k + 1}= (k – 1)2^{k + 1} + 2 + (k + 1)2^{k + 1}

[by equation (i)]

Taking the common terms out

= 2^{k + 1} (k – 1 + k + 1) + 2

so, we get

= 2^{k + 1}.2k + 2

= 2^{(k + 1) + 1}.k + 2

= [(k + 1) – 1]2^{(k + 1) + 1} + 2

∴ P(k + 1) is true, when P(k) is true.

Hence, from the principle of mathematical induction, the statement P(n) is true ∀ n ∈ N.

$$\textbf{9.}\space\frac{\textbf{1}}{\textbf{2}}+\frac{\textbf{1}}{\textbf{4}}+\frac{\textbf{1}}{\textbf{8}}+...+\frac{\textbf{1}}{\textbf{2}^{\textbf{n}}}\\\textbf{1}-\frac{\textbf{1}}{\textbf{2}^{\textbf{n}}}.\\\textbf{Sol.}\space\text{Let P(n):}\space\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{2^{n}}\\1-\frac{1}{2^{n}}$$

for n = 1, we get

$$\text{P(1):}1-\frac{1}{2^{1}}\\1-\frac{1}{2}=\frac{1}{2};$$

P(1) statement is true.

For now, consider that P(n) statement is true for n = k,

$$\text{P(k)}=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{2^{k}}\\1-\frac{1}{2^{k}}\space\text{...(i)}$$

For so let us prove that P(n) is also true for n = k + 1,

$$\text{P(k+1) \space:}\space\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{2^{k}}+\frac{1}{2^{k+1}}$$

$$=1-\frac{1}{2^{k}}+\frac{1}{2^{k+1}}$$

[by (i)]

We can write it as

$$=1-\frac{1}{2^{k}}+\frac{1}{2^{k}.2}$$

Taking the common terms but

$$=1-\frac{1}{2^{k}}\bigg(1-\frac{1}{2}\bigg)$$

so we get

$$=1-\frac{1}{2^{k}}.\frac{1}{2}\\=1-\frac{1}{2^{k+1}}$$

∴ P(k + 1) is true, when P(k) is true.

Hence, from the principle of mathematical induction, the statement P(n) is true ∀ n ∈ N.

$$\textbf{10.}\space\frac{\textbf{1}}{\textbf{2.5}}+\frac{\textbf{1}}{\textbf{5.8}}+\frac{\textbf{1}}{\textbf{8.11}}\textbf{+...+}\frac{\textbf{1}}{\textbf{(3n-1)(3n+2)}}$$

$$\textbf{=}\space\frac{\textbf{n}}{\textbf{6n+4}}.\\\textbf{Sol.}\\\space\text{Let}\space\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{(3n-1)(3n+2)}\\=\frac{n}{6n+4}$$

for n = 1, we get

$$\text{P(1):}\space\frac{1}{6+4}=\frac{1}{10}\\=\frac{1}{2.5};\text{P(1) is statement is true.}$$

For consider P(n) statement is true for n = k

$$\text{P(k)}:\frac{1}{2.5}+\frac{1}{5.8}+...+\frac{1}{(3k-1)(3k+2)}\\=\frac{k}{6k+4}$$

For now, let us prove that P(n) is also true for n = k + 1

$$\text{P(k+1)\space:}\space\frac{1}{2.5}+\frac{1}{5.8}+...+\frac{1}{(3k-1)(3k-2)}\\+\frac{1}{[3(k+1)-1][3(k+1)+2]}\\=\frac{k}{6k+4}+\frac{1}{(3k+2)(3k+5)}$$

[by equation (i)]

By simplification of terms

$$=\frac{k}{2(3k+3)}+\frac{1}{(3k+2)(3k+5)}$$

Taking L.C.M.

$$=\frac{k(3k+5)+2}{2(3k+2)(3k+5)}$$

Further calculation

$$=\frac{3k^{2}+5k+2}{2(3k+2)(3k+5)}\\=\frac{3k^{2}+3k+2k+2}{2(3k+2)(3k+5)}\\=\frac{3k(k+1)+2(k+1)}{2(3k+2)(3k+5)}\\=\frac{(3k+2)(k+1)}{2(3k+2)(3k+5)}\\\text{so, we get}\\=\frac{k+1}{2(3k+5)}\\=\frac{k+1}{6k+10}=\frac{k+1}{6k+6+4}\\=\frac{k+1}{6(k+1)+4}$$

∴ P(k + 1) is true, when P(k) is true.

Hence, from the principle of mathematical induction, the statement P(n) is true ∀ n ∈ N.

$$\textbf{11.}\space\frac{\textbf{1}}{\textbf{1.2.3}}+\frac{\textbf{1}}{\textbf{2.3.4}}+\frac{\textbf{1}}{\textbf{3.4.5}}...+\frac{\textbf{1}}{\textbf{n(n+1)(n+2)}}\\=\frac{\textbf{n(n+3)}}{\textbf{4(n+1)(n+2)}}$$

Sol.

$$\text{Let}\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{n(n+1)(n+2)}\\=\frac{n(n+3)}{4(n+1)(n+2)}$$

For n = 1, we get

$$\text{P(1):}\frac{1(1+3)}{4(1+1)(1+2)}\\=\frac{1×4}{4×2×3}=\frac{1}{2×3}$$

P(1) statement is true.

For consider P(n) statement is true for n = k.

$$\text{P(k):}\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{k(k+1)(k+2)}\\=\frac{k(k+3)}{4(k+1)(k+2)}\space\text{...(i)}\\\text{For new, let us prove that P(n) is}\\\text{P(k+1):}\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\\\frac{1}{k(k+1)(k+2)}\\+\frac{1}{(k+1)[(k+1)+1][(k+1)+2]}\\=\frac{k(k+3)}{4(k+1)(k+2)}+\frac{1}{(k+1)(k+2)(k+3)}$$

[by equation (i)]

Taking L.C.M.

$$=\frac{k(k+3)^{2}+4}{4(k+1)(k+2)(k+3)}$$

Further calculation

$$=\frac{k(k^{2}+6k+9)}{4(k+1)(k+2)(k+3)}\\=\frac{k^{3}+6k^{2}+9k+4}{4(k+1)(k+2)(k+3)}$$

Factorise numerator by remainder theorem, we
get

$$=\frac{(k+1)(k^{2}+5k+4)}{4(k+1)(k+2)(k+3)}\\\text{Factorization}\\=\frac{k^{2}+4k+k+4}{4(k+2)(k+3)}\\=\frac{k(k+4)+1(k+4)}{4(k+2)(k+3)}\\=\frac{(k+1)(k+4)}{4(k+2)(k+3)}\\=\frac{(k+1)[(k+1)+3]}{4[(k+1)+1][(k+1)+2]}$$

∴ P(k + 1) is true, when P(k) is true.

Hence, from the principle of mathematical induction, the statement is true ∀ n ∈ N.

$$\textbf{12. a + ar + ar}^\textbf{2} \textbf{+ … + ar}^\textbf{n – 1}=\frac{\textbf{a(r}^{\textbf{n}}\textbf{-1})}{\textbf{r-1}}.$$

Sol.

$$\text{Let P(n) : a + ar + ar}^{2}+...+\text{ar}^{n-1}\\=\frac{a(r^{n}-1)}{r-1}$$

for n = 1, we get

$$\text{P(1)}:\frac{a(r-1)}{r-1}\\=\frac{a(r-1)}{r-1}=a;\text{(P) statement is true.}$$

for now, consider that P(n) statement is true for n = k,

$$\text{P(k) : a + ar + ar}^{2}+...+ar^{k-1}\\=\frac{a(r^{k}-1)}{r-1}\space\text{...(i)}$$

For P(n) will also be true for n = k + 1,

P(k + 1) : a + ar + ar² + … + ar^{k – 1} + ar^{(k + 1) – 1}

$$=\frac{a(r^{k}-1)}{r-1}+ar^{k}$$

Taking L.C.M.

$$=\frac{a(r^{k}-1)+ar^{k}(r-1)}{r-1}\\=\frac{a(r^{k}-1)+ar^{k+1}-ar^{k}}{r-1}$$

∴ P(k + 1) is true, when P(k) is true.

Hence, from the principle of mathematical induction, the statement P(n) is true ∀ n ∈ N.

$$\textbf{13.}\bigg(\textbf{1+}\frac{\textbf{3}}{\textbf{1}}\bigg)\bigg(\textbf{1+}\frac{\textbf{5}}{\textbf{4}}\bigg)\bigg(\textbf{1+}\frac{\textbf{7}}{\textbf{9}}\bigg)\\...\bigg(\textbf{1+}\frac{\textbf{2n+1}}{\textbf{n}^{\textbf{2}}}\bigg)=\textbf{(n+1)}^{\textbf{2}}.$$

$$\textbf{Sol.}\space\text{Let P(n):}\bigg(1+\frac{3}{1}\bigg)\bigg(1+\frac{5}{4}\bigg)\bigg(1+\frac{7}{9}\bigg)\\...\bigg(1+\frac{2n+1}{n^{2}}\bigg)=(n+1)^{2}$$

for n = 1, we get

$$\text{P(1) : (1 + 1)}^2 = 2^2 = 4 =\bigg(1+\frac{3}{1}\bigg);\\\text{P(1) statement is true.}$$

for consider P(n) statement is true for n = k,

$$\text{P(k):}\bigg(1+\frac{3}{1}\bigg)\bigg(1+\frac{5}{4}\bigg)\\...\bigg(1+\frac{2k+1}{k^{2}}\bigg)=(k+1)^{2}\text{...(i)}$$

for now, P(n) will also be true for n = k + 1,

$$\text{P(k+1)=}\bigg(1+\frac{3}{1}\bigg)\bigg(1+\frac{5}{4}\bigg)\bigg(1+\frac{7}{9}\bigg)\\...\bigg(1+\frac{2k+1}{k^{2}}\bigg)$$

$$\text{P(k+1)=}\bigg(1+\frac{3}{1}\bigg)\bigg(1+\frac{5}{4}\bigg)\bigg(1+\frac{7}{9}\bigg)\\...\bigg(1+\frac{2k+1}{k^{2}}\bigg)\bigg(1+\frac{2(k+1)+1}{(k+1)^{2}}\bigg)\\=(k+1)^{2}\bigg(1+\frac{2k+3}{(k+1)^{2}}\bigg)$$

[by (i)]

Taking L.C.M.

$$=(k+1)^{2}\bigg[\frac{(k+1)^{2}+2k+3}{(k+1)^{2}}\bigg]$$

so, we get

= (k + 1)² + 2k + 3

By further calculation

= k² + 1 + 2k + 2k + 3

= k² + 4k + 4 = (k + 2)²

= [(k + 1) + 1]²

∴ P(k + 1) is true, when P(k) is true.

Hence, from the principle of mathematical induction, the statement P(n) is true ∀ n ∈ N.

$$\textbf{14.}\bigg(\textbf{1}+\frac{\textbf{1}}{\textbf{1}}\bigg)\bigg(\textbf{1}+\frac{\textbf{1}}{\textbf{2}}\bigg)\bigg(\textbf{1}+\frac{\textbf{1}}{\textbf{3}}\bigg)\\...\bigg(\textbf{1}+\frac{\textbf{1}}{\textbf{n}}\bigg)=\textbf{(n+1).}$$

$$\textbf{Sol.}\space\text{P(n):}\bigg(1+\frac{1}{1}\bigg)\bigg(1+\frac{1}{2}\bigg)\bigg(1+\frac{1}{3}\bigg)\\...\bigg(1+\frac{1}{n}\bigg)=\text{(n+1).}$$

for n = 1, we get

$$\text{P(1) : 1 + 1 =}1+\frac{1}{1};\\\text{P(1) statement is true.}$$

for consider that P(n) statement is true for n = k,

$$\text{P(k) : }\bigg(1+\frac{1}{1}\bigg)\bigg(1+\frac{1}{2}\bigg)\\...\bigg(1+\frac{1}{k}\bigg)=(k+1)\space...(i)$$

for now, P(n) is also true for n = k + 1,

$$\text{P(k+1):}\bigg(1+\frac{1}{1}\bigg)\bigg(1+\frac{1}{2}\bigg)\\...\bigg(1+\frac{1}{k}\bigg)\bigg(1+\frac{1}{k+1}\bigg)\\=\text{(k+1)}\bigg(1+\frac{1}{k+1}\bigg)$$

[by equation (i)]

Taking L.C.M.

$$=(k+1)\bigg(\frac{k+1+1}{k+1}\bigg)\\=(k+1)\bigg(\frac{k+2}{k+1}\bigg)$$

By further calculation

= k + 2

= (k + 1) + 1

∴ P(k + 1) is true, when P(k) is true.

Hence, from the principle of mathematical induction, the statement P(n) is true ∀ n ∈ N.

$$\textbf{15. 1}^\textbf{2} + \textbf{3}^\textbf{2} + \textbf{5}^\textbf{2} \textbf{+ … +} \textbf{(2n – 1)}^\textbf{2}\\ =\frac{\text{\textbf{n(2n-1)(2n+1)}}}{\textbf{3}}.$$

Sol. Let P(n) : 1² + 3² + 5² + … + (2n – 1)²

$$=\frac{n(2n-1)(2n+1)}{3}$$

for n = 1, we get

$$\text{P(1) :}\space\frac{1(2×1-1)(2×1+1)}{3}\\=\frac{1×1×3}{3}=1×1=1^{2};\\\text{P(1) statement is true.}$$

for consider, P(n) statement is true for n = k,

P(k) = 1² + 3² + … + (2k – 1)²

$$=\frac{k(2k-1)(2k+1)}{3}\space\text{...(i)}$$

for now, let us prove that P(n) will be true for
n = k + 1,

P(k + 1) = 1² + 3² + 5² + … + (2k – 1)² + [2(k + 1) – 1]

$$=\frac{k(2k-1)(2k+1)}{3}+[2k+1]^{2}$$

{by using equation (i)}

Taking L.C.M.

$$=\frac{k(2k-1)(2k+1)+3(2k+1)^{2}}{3}$$

Taking the common terms out

$$=\frac{(2k+1)[k(2k-1)+3(2k+1)]}{3}$$

By further simplication

$$=\frac{(2k+1)(2k^{2}-k+6k+3)}{3}\\\text{so, we get}\\=\frac{(2k+1)(2k^{2}+5k+3)}{3}\\=\frac{(2k+1)(2k^{2}+2k+3k+3)}{3}\\=\frac{(2k+1)[2k(k+1)+3(k+1)]}{3}$$

splitting the terms

$$=\frac{(2k+1)(k+1)(2k+3)}{3}\\\text{Now, we get}\\=\frac{(k+1)(2k+2-1)(2k+2+1)}{3}\\=\frac{(k+1)[2(k+1)-1][2(k+1)+1]}{3}$$

∴ P(k + 1) is true, when, P(k) is true.

Hence, from the principle of mathematical induction, the statement P(n) is true ∀ n ∈ N.

$$\textbf{16.}\space\frac{\textbf{1}}{\textbf{1.4}}+\frac{\textbf{1}}{\textbf{1.6}}+\frac{\textbf{1}}{\textbf{7.10}}\textbf{+...+}\frac{\textbf{1}}{\textbf{(3n-2)(3n+1)}}\\=\frac{\textbf{n}}{\textbf{3n+1}}.$$

$$\textbf{Sol.}\space\text{Let P(n):\space}\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...\\\ +\frac{1}{(3n-2)(3n+1)}=\frac{n}{3n+1}.$$

for n = 1, we get

$$\text{P(1)\space:\space}\frac{1}{3×1+1}=\frac{1}{4}=\frac{1}{1.4};\\\text{P(1) statement is true.}$$

for consider P(n) statement is true for n = k,

$$\text{P(k)}\space:\space\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...\\+\frac{1}{(3k-2)(3k+1)}=\frac{k}{3k+1}\space\text{...(i)}$$

For now, let us prove P(n) is also true for n = k + 1,

$$\text{P(k+1)\space:\space}\frac{1}{1.4}+\frac{1}{4.7}+...\\+\frac{1}{(3k-2)(3k+1)}+\frac{1}{[3(k+1)-2][3(k+1)+1]}$$

$$=\frac{k}{3k+1}+\frac{1}{(3k+1)(3k+4)}\\\lbrace\text{by equation (i)}\rbrace\\=\frac{k(3k+4)+1}{(3k+1)(3k+4)}=\frac{3k^{2}+4k+1}{(3k+1)(3k+4)}\\=\frac{3k^{2}+3k+k+1}{(3k+1)(3k+4)}=\frac{3k(k+1)+1(k+1)}{(3k+1)(3k+4)}\\=\frac{(3k+1)(k+1)}{(3k+1)(3k+4)}=\frac{k+1}{3k+4}$$

At last, we get

$$=\frac{k+1}{3k+3+1}=\frac{k+1}{3(k+1)+1}$$

∴ P(k + 1) is true when P(k) is true.

Hence, from the principle of mathematical induction, the statement P(n) is true ∀ n ∈ N.

$$\textbf{17.\space}\frac{\textbf{1}}{\textbf{3.5}}+\frac{\textbf{1}}{\textbf{5.7}}+\frac{\textbf{1}}{\textbf{7.9}}+...\\+\frac{\textbf{1}}{\textbf{(2n+1)(2n+3)}}=\frac{\textbf{n}}{\textbf{3(2n+3).}}\\\textbf{Sol.}\space\text{Let P(n):}\space\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...\\ + \frac{1}{(2n+1)(2n+3)}=\frac{n}{3(2n+3)}$$

for n = 1, we get

$$\text{P(1)\space:\space}\frac{1}{3(2×1+3)}=\frac{1}{3.5};\\\text{P(1) statement is true.}$$

For consider, P(n) statement is true for

$$\text{P(k)\space:\space}\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{(2k+1)(2k+3)}\\=\frac{k}{3(2k+3)}\space\text{...(i)}$$

For now, we’ll prove that P(n) is also true for n = k + 1,

$$\text{P(K+1)\space:\space}\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...\\+\frac{1}{(2k+1)(2k+3)}\\+\frac{1}{[2(k+1)+1][2(k+1)+3]}\\=\frac{k}{3(2k+3)}+\frac{1}{(2k+3)(2k+5)}\\\text{[by\space(i)]}$$

Taking L.C.M.

$$=\frac{k(2k+5)+3}{3(2k+3)(2k+5)}\\=\frac{2k^{2}+5k+3}{3(2k+3)(2k+5)}=\frac{2k^{2}+2k+3k+3}{3(2k+3)(2k+5)}\\=\frac{2k(k+1)+3(k+1)}{3(2k+3)(2k+5)}=\frac{(k+1)(2k+3)}{3(2k+3)(2k+5)}\\=\frac{k+1}{3(2k+5)}=\frac{k+1}{3(2k+2+3)}\\\text{At last, we get}\\\frac{k+1}{3[2(k+1)+3]}$$

∴ P(k + 1) is true when P(k) is true.

Hence, from the principle of mathematical induction. The statement P(n) is true ∀ n ∈ N.

$$\textbf{18. 1 + 2 + 3 + … + n}\lt\frac{\textbf{1}}{\textbf{8}}(\textbf{2n+1)}^{\textbf{2}}.$$

$$\textbf{Sol.}\\\space\text{Let P(n) : 1 + 2 + 3 + … + n}\lt\frac{1}{8}(2n+1)^{2}$$

for n = 1, we get

$$\text{P(1)\space:\space1}\lt\frac{1}{8}(2×1+1)^{2}\\\Rarr\space1\lt\frac{1}{8}×3^{2}\\\Rarr\space1\lt\frac{9}{8};\\\text{P(1) statement is true}$$

For considering P(n) statement is true for n = k,

$$\text{P(k) : 1 + 2 + 3 + … }\\+\space k\lt\frac{1}{8}(2k+1)^{2}\space\text{...(i)}$$

For now, let us prove that P(n) is also true for
n = k + 1,

P(k + 1) : 1 + 2 + 3 + … + k + k + 1

$$\lt\frac{1}{8}(2k+1)^{2}+(k+1)[\text{by using (i)}]$$

Expanding terms using formula {(a + b)² = a²+ b² + 2ab}

$$\lt\frac{4k^{2}+1+4k+8k+8}{8}\\\text{By further calculation}\\\lt\frac{4k^{2}+12k+9}{8}\\\text{so we get}\\\lt\frac{(2k+3)^{2}}{8}=\frac{(2k+2+1)^{2}}{8}\\\lt\frac{[2(k+1)+1]^{2}}{8}$$

∴ P(k + 1) is true when P(k) is true.

Hence, from the principle of mathematical induction the statement P(n) is true ∀ n ∈ N.

19. n(n + 1) (n + 5) is a multiple of 3.

Sol. Let P(n) : n(n + 1) (n + 5) is a multiple of 3.

for n = 1, we get

P(1) : 1(1 + 1) (1 + 5) = 1 × 2 × 6

= 12 = 3 × 4

which is a multiple of 3 i.e., true.

For consider P(n) statement is true for n = k,

P(k) : k(k + 1) (k + 5) = 3λ; which is true.

⇒ k(k² + 5k + k + 5) = 3λ

⇒ k³ + 6k² + 5k = 3λ …(i)