10. (2n -1)2

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Solution

The n th term of the series is given as,

a n = ( 2 n −1 ) 2 =4 n 2 −4n+1

So, the sum of series is,

S n = ∑ k=1 n a k = ∑ k=1 n ( 4 k 2 −4k+1 ) =4 ∑ k=1 n k 2 −4 ∑ k=1 n k + ∑ k=1 n 1 = 4n( n+1 )( 2n+1 ) 6 − 4n( n+1 ) 2 +n

Solve further,

S n =n[ 2( 2 n 2 +3n+1 ) 3 −2( n+1 )+1 ] =n[ 4 n 2 +6n+2−6n−6+3 3 ] =n[ 4 n 2 −1 3 ] = n( 2n+1 )( 2n−1 ) 3

Therefore, the sum of n terms of series is n( 2n+1 )( 2n−1 ) 3 .

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