If S1, S2, S3 are the sum of first n natural numbers, their squares and their cubes, respectively, show that 9S22 = S3 (1 + 8S1).

Asked by Pragya Singh | 1 year ago |  66

##### Solution :-

According to the conditions given in the question,

$$s_1=\dfrac{n(n+1)}{2}$$

$$s_3=\dfrac{n^2(n+1)^2}{4}$$

Therefore,

$$s_3(1+8s_1)=\dfrac{n^2(n+1)^2}{4}$$

$$[1+\dfrac{8n(n+1)}{2}]$$

$$\dfrac{n^2(n+1)^2}{4}[1+4n^2+4]$$

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

$$\dfrac{[n(n+1)(2n+1)]^2}{4}$$ .........(1)

And.

$$9s^2_2=9 \dfrac{[n(n+1)(2n+1)]^2}{(6)^2}$$

$$\dfrac{9}{36}[n+(n+1)(2n+1)]^2$$

$$= \dfrac{[n(n+1)(2n+1)]^2}{4}$$ .............(2)

Therefore, we get

$$9s^2_2= s_3(1+8s_1)$$ from (1) and (2).

Answered by Abhisek | 1 year ago

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