Question

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

Solution

## 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 . MathematicsMath - NCERTStandard XI

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