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JEE Main · 2019 · 09 Apr 2019 · Shift 2 · Sequences and Series

The sum of the series

1 + 2 \times 3 + 3 \times 5 + 4 \times 7 + \dots

upto

1 1^{th}

term is

(A)915
(B)946✓ Correct
(C)945
(D)916

Answer: (B)

Solution

The products in the series are

1 \cdot 1, 2 \cdot 3, 3 \cdot 5, 4 \cdot 7, \dots

, where the second factors are the odd numbers, so the

n

th term is

n (2 n - 1) = 2 n^{2} - n

. Summing to

n = 11

,

n = 1 \sum 11 (2 n^{2} - n) = 2 \cdot \frac{11 \cdot 12 \cdot 23}{6} - \frac{11 \cdot 12}{2} = 2 \cdot 506 - 66 = 1012 - 66 = 946

.

JEE Main · 2019 · 09 Apr 2019 · Shift 2 · Single correct option

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