Question

Find the sum of the following series up to n terms:

Answer 1

The given sequence is 1 3 1 + 1 3 + 2 3 1+3 + 1 3 + 2 3 + 3 3 1+3+5 +…

The n th term of the sequence a n is given by,

a n = 1 3 + 2 3 + 3 3 +…+ n 3 1+3+5+…+( 2n−1 ) = [ n( n+1 ) 2 ] 2 1+3+5+…+( 2n−1 ) (1)

In the above expression, the sequence in the denominator 1+3+5+…+( 2n−1 ) is in an A.P.

Here,

a=1 d=2

The sum of the above sequence is given by,

1+3+5+…+( 2n−1 )= n 2 [ 2×1+( n−1 )×2 ] =[ n+ n 2 −n ] = n 2

Substitute the value of 1+3+5+…+( 2n−1 ) in equation (1).

a n = [ n( n+1 ) 2 ] 2 n 2 = n 2 ( n+1 ) 2 4 n 2 = ( n+1 ) 2 4 = 1 4 ( n 2 +1+2n ) (2)

The sum of equation (2) is given by,

S n = ∑ k=1 n a k

Substitute the value of a k in the above expression from equation (2).

S n = ∑ k=1 n ( 1 4 ( k 2 +1+2k ) ) = ∑ k=1 n ( 1 4 k 2 + 1 4 + 1 4 ×2k ) = ∑ k=1 n 1 4 k 2 + ∑ k=1 n 1 4 ×2k+ ∑ k=1 n 1 = 1 4 ∑ k=1 n k 2 + 1 2 ∑ k=1 n k+ ∑ k=1 n 1

Further simplify the above expression.

S n = 1 4 n( n+1 )( 2n+1 ) 6 + 1 2 n( n+1 ) 2 + 1 4 n = n[ ( n+1 )( 2n+1 )+6( n+1 )+6 ] 24 = n[ 2 n 2 +3n+1+6n+6+6 ] 24 = n[ 2 n 2 +9n+13 ] 24

Thus, the sum of the sequence 1 3 1 + 1 3 + 2 3 1+3 + 1 3 + 2 3 + 3 3 1+3+5 +… is n[ 2 n 2 +9n+13 ] 24 .