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Sigma n

If i - 1n∑ i ...

Question

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

A

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

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B

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

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C

n(3n + 2)

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D

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

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Solution

The correct option is A $\frac{n (3 n - 1)}{2}$
$n \sum i - 1 = 3 n \sum i - 1 i - 2 n \sum i - 1 = 3 \frac{n (n + 1)}{2} - 2 n = \frac{n (3 n - 1)}{2} .$

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

(1 + i) (1 + 2 i) (1 + 3 i) . . . . (1 + n i) = α + i β

2.5.10 . . . (1 + n^{2}) =

(1 - i) (1 + 2 i) (1 + 3 i) . . . . (1 + n i) = α + i β

2 \cdot 5 \cdot 10 . . . . (1 + n^{2}) =

z = \frac{(1 + i) (1 + 2 i) (1 + 3 i) \dots \dots \dots (1 + n i)}{(1 - i) (2 - i) (3 - i) \dots \dots \dots (n - i)}

i = \sqrt{- 1}, n \in N

, then principal argument of

z

(1 + i) (1 + 2 i) (1 + 3 i) . . . . (1 + n i) = x + i y

2.5.10....... (1 + n^{2}) = x^{2} + y^{2}

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