If π ∑ i = 1 i = n ( n + 1 ) 2 , then π ∑ i = 1 ( 3 i − 2 ) = Options: A . &nbsp n ( 3 n − 1 ) 2 B . &nbsp n ( 3 n + 1 ) 2 C . &nbsp n(3n + 2) D . &nbsp n ( 3 n + 1 ) 4

Answer: Option A : A π ∑ i = 1 = 3 π ∑ i = 1 i - 2 π ∑ i = 1 1 = 3 n ( n + 1 ) 2 - 2n = n ( 3 n − 1 ) 2

If π∑i=1i = n(n+1)2, then π∑i=1(3iȡ

Question -

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

Options:

A .

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

B .

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

C . n(3n + 2)

D .

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

Answer: Option A
:
A

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

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