Principle of Mathematical Induction

17. We can write the given statement as:-

$\frac{1}{3\cdot 5}+\frac{1}{5\cdot 7}+\frac{1}{7\cdot 9}+?+\frac{1}{(2n+1)(2n+3)}=\frac{n}{3(2n+3)}$

For n = 1,

We get $P(1)=\frac{1}{3\cdot 5}=\frac{1}{15}=\frac{1}{3(2\cdot 1+3)}=\frac{1}{3(2+3)}=\frac{1}{3*5}=\frac{1}{15}$

Which is true.

Consider P(k) be true for some positive integer k.

$P(K)=\frac{1}{3\cdot 5}+\frac{1}{5\cdot 7}+\frac{1}{7\cdot 9}+\dots +\frac{1}{(2k+1)(2k+3)}=\frac{k}{3(2k+3)}$ (1)

Now, let us prove that P(k+ 1) is true.

Now,

P(k +1) = $1$ $\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{(2k+1)(2k+3)}+\frac{1}{[2(k+1)+1][2(k+1)+3]}$

By using (1),

$\begin{array}{l}=\frac{k}{3(2k+3)}+\frac{1}{(2k+2+1)(2k+2+3)}\\ =\frac{k}{3(2k+3)}+\frac{1}{(2k+3)(2k+5)}\\ =\frac{1}{(2k+3)}\left[\frac{k}{3}+\frac{1}{(2k+5)}\right]\\ =\frac{1}{(2k+3)}\left[\frac{k(2k+5)+3}{3(2k+5)}\right]\end{array}$

$=\frac{1}{(2k+3)}\left[\frac{2k^2+5k+3}{3(2k+5)}\right]$

= $\frac{1}{2k+3}\left[\frac{2k^2+2k+3k+3}{3(2k+5)}\right]$

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

$\begin{array}{l}=\frac{1}{(2k+3)}\frac{(2k+3(k+1)}{\cdot (2k+1)\cdot 3}\\ =\frac{k+1}{3(2k+5)}\\ =\frac{(k+1)}{3\{2(k+1)+3\}}\end{array}$

P(k+ 1) is true wheneverP(k) is true.

Therefore, from the principle of mathematical induction, theP(n) is true for all natural number n.

16. Let the given statement as

P(n)= $\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}$ + … + $\frac{1}{(3n-2)(3n+1)}=\frac{n}{(3n+1)}$

If n=1, then

P(1)= $\frac{1}{1.4}$ = $\frac{1}{4}$ = $\frac{1}{(3.1+1)}$ = $\frac{1}{4}$

which is true.

Consider P(k)be true for some positive integer k

P(k)= $\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}$ + … + $\frac{1}{(3k-2)(3k+1)}$ = $\frac{k}{(3k+1)}$ ------------------(1)

Now, let us prove P(k+1) is true.

P(k+1)= $\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}$ + … + $\frac{1}{(3k-2)(3k+1)}+\frac{1}{[3(k+1)-2][3(k+1)+1]}$

By using (1),

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

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

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

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

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

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

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

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

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

? P(k+1) is true whenever P(k) is true.

Therefore, from the principle of mathematical induction, the P(n) is true for all natural number n.

15. We can write the given statement as

P(n)=1²+3²+5²+ … + (2n – 1)²= $\frac{n(2n-1)(2n+1)}{3}$

forn=1

P(1)=1²=1= $\frac{1(2.1-1)(2.1+1)}{3}$

= $\frac{1(1)(3)}{3}=1$ which is true.

Consider P(k) be true for some positive integer k

P(k)=1²+3²+5²+ … + (2n – 1)²= $\frac{k(2k-1)(2k+1)}{3}$ ------------------(1)

Now, let us prove that P(k+1) is true.

Here,

1²+3²+5²+ … +(2k – 1)²+(2(k+1) –1)²

By using (1),

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

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

= $\frac{(2k+1)[k(2k-1)+3(2k+1)]}{3}$

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

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

we can write as,

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

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

= $\frac{(2k+1)(2k+3)(k+1)}{3}$

= $\frac{(2k+1)(k+1)(2k+3)}{3}$

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

P(k+1) is true whenever P(k) is true.

Hence, from the principle of mathematical induction, the P(n) is true for all natural number n.

14. Let the given statement be P(n) i.e.,

P(n)= $\left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)$ … $\left(1+\frac{1}{n}\right)=(n+1)$

If n =1

P(1)= $\left(1+\frac{1}{1}\right)$ = 2 =1+1= 2

which is true.

Assume that P(k) is true for some positive integer k i.e.,

P(k): $\left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)$ … $\left(1+\frac{1}{k}\right)=(k+1)$ .---------------------(1)

Now, let us prove that P(k+1) is true.

Here,

P(k+1)= $\left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)$ … $\left(1+\frac{1}{k}\right)+\left(1+\frac{1}{(k+1)}\right)$

By using (1), we get

(k+1). $\left(1+\frac{1}{k+1}\right)$

L.C.M.=(k+1). $\left(\frac{k+1+1}{k+1}\right)$

= (k+1)+1

? P(k+1) is true whenever P(k) is true.

Therefore from the principle of mathematical induction the P(n) is true for all natural numbers n.

13. We can write given statement as

P(n): $\left(1+\frac{3}{1}\right)\left(1+\frac{5}{4}\right)\left(1+\frac{7}{9}\right)$ … $\left(1+\frac{(2n+1)}{n^2}\right)={(n+1)}^2$

If n=1, we get

P(1): $\left(1+\frac{3}{1}\right)$ =4=(1+ 1)²=2²=4

which is true.

Consider P(k) be true for some positive integer k.

$\left(1+\frac{3}{1}\right)\left(1+\frac{5}{4}\right)\left(1+\frac{7}{9}\right)$ … $\left(1+\frac{(2k+1)}{k^2}\right)={(k+1)}^2$ (1)

Now, let us prove that P(k+1) is true.

$\left(1+\frac{3}{1}\right)\left(1+\frac{5}{4}\right)\left(1+\frac{7}{9}\right)$ … $\left(1+\frac{(2k+1)}{k^2}\right)+\left(1+\frac{(2(k+1)+1)}{{(k+1)}^2}\right)$

By using (1)

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

=(k+1)² $\left[\frac{{(k+1)}^2+2(k+1)+1}{{(k+1)}^2}\right]$

=(k+1)²+2(k+1)+1

={(k+1)+1}²

P(k+1) is true whenever P(k) is true.

Therefore, by principle of mathematical induction, the P(n) is true for all natural number n.

12. Let the given statement be P(n) i.e.,

P(n)=a+ar+ar²+ … +ar^n-1== $\frac{a\left(r^n-1\right)}{r-1}$

If n = 1, we get

P(1)=a= $\frac{a\left(r^1-1\right)}{r-1}$ =a

which is true.

Consider P(k) be true for some positive integer k

a+ar+ar²+ … +ar^k-1= $\frac{a\left(r^k-1\right)}{r-1}$ (1)

Now, let us prove that P(k+1) is true.

Here, {a+ar+ar²+ … +ar^k-1}+ar^{(k+1) –1}

By using (1),

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

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

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

= $\frac{a{r}^{k+1}-a}{r-1}$

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

P(k+1) is true whenever P(k) is true.

Therefore, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e.,

11. we can write the given statement as

$P(n)=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}$ + … + $\frac{1}{n(n+1)(n+2)}$ = $\frac{n(n+3)}{4(n+1)(n+2)}$

If n=1,

P(1)= $\frac{1}{1.2.3}$ = $\frac{1}{6}$ = $\frac{1(1+3)}{4(1+1)(1+2)}$ = $\frac{4}{4*2*3}$ = $\frac{1}{6}$

which is true.

Consider P(k) be true for some positive integer k

$\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}$ + … + $\frac{1}{k(k+1)(k+2)}$ = $\frac{k(k+3)}{4(k+1)\left(k+2\right)}$

Let us prove that P(k+1) is true,

$\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}$ + … + $\frac{1}{k(k+1)(k+2)}+\frac{1}{(k+1)(k+2)(k+3)}$ .

By equation (1), we get

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

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

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

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

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

= $\frac{1}{(k+1)(k+2)}\left\{\frac{k^3+2k^2+k+4k^2+8k+4}{4(k+3)}\right\}$

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

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