An odd square i.e ${[2 n + 1]}^{2}$ is equal to sum of n terms of an A.P. in increased by unity.Find the common difference of the A.P?

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Solution

$S_{n} + 1 = {(2 n + 1)}^{2}$ as above
$T_{n} = S_{n} - S_{n - 1} = 4 n^{2} + 4 n + 1 - 4 n^{2} + 4 n - 1 = 8 n$
$T = 8, 16, 24, . . . . .$
So the common difference is $8$

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An odd square i.e [2n+1]2 is equal to sum of n terms of an A.P. in increased by unity.Find the common difference of the A.P?

Sn+1=(2n+1)2 as aboveTn=Sn−Sn−1=4n2+4n+1−4n2+4n−1=8nT=8,16,24,.....So the common difference is 8

An odd square i.e ${[2 n + 1]}^{2}$ is equal to sum of n terms of an A.P. in increased by unity.Find the common difference of the A.P?

$S_{n} + 1 = {(2 n + 1)}^{2}$ as above
$T_{n} = S_{n} - S_{n - 1} = 4 n^{2} + 4 n + 1 - 4 n^{2} + 4 n - 1 = 8 n$
$T = 8, 16, 24, . . . . .$
So the common difference is $8$