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

$$\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+...$$

$$+\dfrac{1}{(3n-1)(3n+2)}=\dfrac{n}{(6n+4)}$$

Asked by Abhisek | 2 years ago |  71

##### Solution :-

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

$$P(n): \dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+...$$

$$+\dfrac{1}{(3n-1)(3n+2)}=\dfrac{n}{(6n+4)}$$

For n=1,

L.H.S. = $$\dfrac{1}{2.5}=\dfrac{1}{10}$$

R.H.S.= $$\dfrac{1}{6.1+4}=\dfrac{1}{10}$$

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

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

$$\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+...$$

$$+\dfrac{1}{(3k+1)(3k+2)}=\dfrac{k}{(6k+4)}$$.....(i)

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

Consider

$$\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+...$$

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

$$\dfrac{k}{6k+4}+ \dfrac{1}{(3k+3-1)(3k+3+2)}$$

$$\dfrac{k}{6k+4}+ \dfrac{1}{(3k+2)(3k+5)}$$

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

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

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

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

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

$$\dfrac{(k+1)}{6(k+1)+4}$$

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 Pragya Singh | 2 years ago

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