CAT 2018 Question Paper Shift 2 Solved Paper

Question : 72 of 100

Marks: +1, -0

The value of the sum 7 x 11 + 11 x 15 + 15 x 19 + ...+ 95 x 99 is
Options

Solution:

\N^{\th}

term of the series can be written as

{tn}=(4 {n}+3)(4 {n}+7)

=16 n^{2}+40 n+21

Σ tn=16 Σ n^{2}+40 Σ n+21 Σ 1

=16 \;{ {n}( {n}+1)(2 {n}+1)}/{6}+40 \;{ {n}( {n}+1)}/{2}+2\ln

here

{n}=23(7,11,15 . . . . .95

is an AP with common different 4 with 23 terms)

∑ t_{n}=\;{16 × 23 × 24 × 47}/{6}+20 × 23 × 24+21 × 23

=80707

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