Prove the following by using the principle of mathematical induction for all $$n\in N$$

$$\dfrac{1}{3.5}+ \dfrac{1}{5.7}+ \dfrac{1}{7.9}+...+ \dfrac{1}{(2n+1)(2n+3)}= \dfrac{n}{2(2n+3)}$$

Asked by Pragya Singh | 1 year ago |  65

##### Solution :-

Let us denote the given equality by P(n), i.e.,

$$P(n): \dfrac{1}{3.5}+ \dfrac{1}{5.7}+ \dfrac{1}{7.9}+...$$

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

For n=1,

L.H.S. = $$\dfrac{1}{3.5}$$

R.H.S.= $$\dfrac{1}{3(2.1+3)}=\dfrac{1}{3.5}$$

Therefore,P(n) is true for n=1.

Let us assume that P(k) is true for some positive integer k , i.e.,

$$P(k): \dfrac{1}{3.5}+ \dfrac{1}{5.7}+ \dfrac{1}{7.9}+...$$

$$+ \dfrac{1}{(2k+1)(2k+3)}= \dfrac{k}{2(2k+3)}$$ ........(i)

Now, we have to prove that P(k+1) is also true.

Consider

$$[\dfrac{1}{3.5}+ \dfrac{1}{5.7}+ \dfrac{1}{7.9}+...$$

$$+ \dfrac{1}{(2k+1)(2k+3)}]+$$

$$+ \dfrac{1}{\{2(k+1)+1\}\{2(k+1)+3\}}$$

$$=\dfrac{k}{3(2k+3)}+ \dfrac{1}{(2k+3)(2k+5)}$$......... [Using (i)]

$$\dfrac{1}{(2k+3)}[\dfrac{k}{3}+\dfrac{1}{(2k+5)}]$$

$$\dfrac{1}{(2k+3)}[\dfrac{k(2k+5)+3}{3(2k+5)}]$$

$$\dfrac{1}{(2k+3)}[\dfrac{2k^2+5k+3}{3(2k+5)}]$$

$$\dfrac{1}{(2k+3)}[\dfrac{2k^2+2k+3k+3}{3(2k+5)}]$$

$$\dfrac{1}{(2k+3)}[\dfrac{2k(k+1)+3(k+1)}{3(2k+5)}]$$

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

$$\dfrac{(k+1)}{3\{2(k+1)+3\}}$$

Therefore, P(k+1) holds whenever P(k) holds.

Hence, the given equality is true for all natural numbers i.e., N by the principle of mathematical induction.

Answered by Abhisek | 1 year ago

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