-i-1n 6 i2-a n3-b n2-c n- dn- Then d-A- 3B- 6C- 1D- 0

The correct option is D 0 ∑_i=1^n 6i^2 = 6∑_i=1^n i^2 = 6n/6 (n+1)(2n+1) = n(2n^2 + 3n + 1) = 2n^3 + 3n^2 + n ...

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Summation by Sigma Method

∑i=1n 6 i2=a ...

Question

$\sum_{i = 1}^{n} 6 i^{2} = a n^{3} + b n^{2} + c n + d n$ .Then d =

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Similar questions

$\sum_{i = 1}^{n} 6 i^{2} = a n^{3} + b n^{2} + c n + d$ . Then

n \sum s = 1 {s \sum r = 1 r} = a n^{3} + b n^{2} + c n

n \sum r = 1 T_{r} = 5 n + 2

n \sum r = 1 T_{r}^{2} = a n^{3} + b n^{2} + c n + d

d + b + c

n^{t h}

(T_{n})

a n^{3} + b n^{2} + c n + d

T_{n}

a, b, c, d

(a^{n} + b^{n}), (b^{n} + c^{n}), (c^{n} + d^{n})

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