Sequences And Series Question 17
Question 17 - 31 January - Shift 2
The sum $1^{2}-2.3^{2}+3.5^{2}-4.7^{2}+5.9^{2}-\ldots \ldots . .+15.29^{2}$ is
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Answer: 6952
Solution:
Formula: General term ( $n^{\text {th }}$ term) of an A.G.P.
Separating odd placed and even placed terms, we get
$S=(1.1^{2}+3.5^{2}+\ldots .15 .(29)^{2})-(2.3^{2}+4.7^{2}.$
$+\ldots .+14 .(27)^{2}$
$S=\sum _{n=1}^{8}(2 n-1)(4 n-3)^{2}-\sum _{n=1}^{7}(2 n)(4 n-1)^{2}$
Applying summation formula we get,
$=29856-22904=6952$
