Question

7. 12(12(122232) +

Answer 1

The given series is 1 2 +( 1 2 + 2 2 )+( 1 2 + 2 2 + 3 2 )........... n th term.

Now, the n th term of the given series is,

a n =( 1 2 + 2 2 + 3 2 +.........+ n 2 ) = n( n+1 )( 2n+1 ) 6 = n( 2 n 2 +3n+1 ) 6 = 2 n 3 +3 n 2 +n 6

Solve further.

a n = 1 3 n 3 + 1 2 n 2 + 1 6 n

So, the sum of series is,

S n = ∑ k=1 n a k = 1 3 ∑ k=1 n k 3 + 1 2 ∑ k=1 n k 2 + 1 6 ∑ k=1 n k = 1 3 [ n( n+1 ) ( 2 ) 3 ] 2 + 1 2 × n( n+1 )( 2n+1 ) 6 + 1 6 × n( n+1 ) 2 = n( n+1 ) 6 +[ n( n+1 ) 2 + ( 2n+1 ) 2 + 1 2 ]

Solve further.

S n = n( n+1 ) 6 [ n 2 +n+2n+2 2 ] = n( n+1 ) 6 [ n( n+1 )+2( n+1 ) 2 ] = n( n+1 ) 6 [ ( n+1 )( n+2 ) 2 ] = n ( n+1 ) 2 ( n+2 ) 12

Thus, the sum of series 1 2 +( 1 2 + 2 2 )+( 1 2 + 2 2 + 3 2 )........... n th termis n ( n+1 ) 2 ( n+2 ) 12 .