If -i-1n i-nn-1-2- then -i-1n3 i-2-A- n3 n-1-2B- n 3 n -1-2C- n3 n-2D- n3 n-1-4

The correct option is A n(3n 1)/2i 1n∑ = 3 i 1n∑ i 2 i 1n∑ = 3 n(n+1)/2 2n = n(3n 1)/2. ...

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Sigma n

If ∑i-1n i=nn...

Question

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

\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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Solution

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Similar questions

I f 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) =

FInd the $n^{t h}$ term of the series $(1)^{3}$ + ( $1^{3}$ + $2^{3}$ ) + ( $1^{3}$ + $2^{3}$ + $3^{3}$ ) + ------------------- n terms.

1, z_{1}, z_{2}, . . . z_{n - 1}

\frac{1}{3 - z_{1}} + \frac{1}{3 - z_{2}} + . . . \frac{1}{3 - z_{n - 1}}

1.2 + 2.3 + 3.4 + \dots + n (n + 1) = \frac{n}{3} (n + 1) (n + 2)

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