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  • Let <img alt="\mathrm{S}_{\mathrm{K}} \frac{1+2+\ldots+\mathrm{K}}{\mathrm{K}}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7BS%7D_%7B%5Cmathrm%7BK%7D%7D%20%5Cfrac%7B1&plus;2

Let \mathrm{S}_{\mathrm{K}} \frac{1+2+\ldots+\mathrm{K}}{\mathrm{K}} and  \sum_{j=1}^n S_j^2=\frac{n}{A}\left(\mathrm{Bn}^2+C n+D\right) , where \mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D} \in \mathrm{N} and A has least value. Then

Option: 1

A + B is divisible by D


Option: 2

A + B = 5 (D –C)


Option: 3

A + C + D is not divisible by B


Option: 4

A + B + D is divisible by 5


Answers (1)

best_answer

\begin{aligned} & S_k=\frac{k+1}{2} \\ & S_k^2=\frac{k^2+1+2 k}{4} \\ & \therefore \sum_{j-1}^n S_j^2=\frac{1}{4}\left[\frac{n(n+1)(2 n+1)}{6}+n+n(n+1)\right] \\ & =\frac{n}{4}\left[\frac{(n+1)(2 n+1)}{6}+1+n+1\right] \\ & =\frac{n}{4}\left[\frac{2 n^2+3 n+1}{6}+n+2\right] \\ & =\frac{n}{4}\left[\frac{2 n^2+9 n+13}{6}\right]=\frac{n}{24}\left[2 n^2+9 n+13\right] \\ & A=24, B=2, C=9, D=13 \end{aligned}

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Divya Prakash Singh

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