Prove the following by using the principle of mathematical induction for all $$n\in N$$
$$\dfrac{1}{1.4}+ \dfrac{1}{4.7}+ \dfrac{1}{7.10}+...+\dfrac{1}{(3n-2)(3n+1)}=\dfrac{n}{(3n+1)}$$

Asked by Pragya Singh | 1 year ago |  60

##### Solution :-

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

$$\dfrac{1}{1.4}+ \dfrac{1}{4.7}+ \dfrac{1}{7.10}+...$$

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

For n=1,

L.H.S.= $$\dfrac{1}{1.4}$$$$\dfrac{1}{4}$$

R.H.S.= $$\dfrac{1}{3.1+1}= \dfrac{1}{4}$$

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

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

$$\dfrac{1}{1.4}+ \dfrac{1}{4.7}+ \dfrac{1}{7.10}+...$$

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

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

Consider

$$\{ \dfrac{1}{1.4}+ \dfrac{1}{4.7}+ \dfrac{1}{7.10}+...$$

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

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

$$\dfrac{k}{3k+1} +\dfrac{1}{(3k+1)(3k+14)}$$ [Using (i)]

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

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

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

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

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

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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