If -i-1- i - nn-1-2- then -i-1- 3i - 2 -

∑_i=1^π = 3∑_i=1^π i 2 ∑_i=1^π 1 = 3 n(n+1)/2 2n = n(3n 1)/2 ...

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Sigma n

If ∑i=1π i = ...

Question

If $π \sum i = 1 i$ = $\frac{n (n + 1)}{2}$ , then $π \sum i = 1 (3 i - 2)$ =

$\frac{n (3 n - 1)}{2}$

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$\frac{n (3 n + 1)}{2}$

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n(3n + 2)

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$\frac{n (3 n + 1)}{4}$

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Similar questions

If $π \sum i = 1 i$ = $\frac{n (n + 1)}{2}$ , then $π \sum i = 1 (3 i - 2)$ =

If $π \sum i = 1 i$ = $\frac{i (i + 1)}{2}$ , then $n \sum i = 1 (3 i - 2)$ =

n \sum i - 1 i = \frac{n (n + 1)}{2}, t h e n n \sum i - 1 (3 i - 2) =

n \sum i - 1 i = \frac{n (n + 1)}{2}, t h e n n \sum i - 1 (3 i - 2) =

n \sum i - 1 i = \frac{n (n + 1)}{2}, t h e n n \sum i - 1 (3 i - 2) =

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