If -i-1- i-ii-1-2- then -i-1n3 i-2-

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

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Sigma n2

If ∑i=1π i=ii...

Question

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

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Solution

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

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) =

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

z_{1} = 2 - i, z_{2} = - 2 + i,

(\frac{z_{1} z_{2}}{z_{1}})

(\frac{1}{z_{1} z_{1}})

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