Question

Question 25 Find the sum of the following series up to n terms: (1^3 / 1) + [(1^3 + 2^3) / (1+3)] + [(1^3 + 2^3 + 3^3) / (1+3+5)] + ...

Class X1 - Maths -Sequences and Series Page 200

Answer 1

1³/1 + (1³ + 2³)/(1 + 3) + ( 1³ + 2³ + 3³)/(1 + 3 + 5)+.....

Tₙ = ( nth term of numerator )/( nth term of denominator )
= (1³ + 2³ + 3³ + ...... n terms )/(1 + 3 + 5 + .... n terms )

nth term of numerator = 1³ + 2³ + 3³ + ..... n terms
= ∑n³ = [n(n + 1)/2]²
nth term of denominator = 1 + 3 + 5 + ...... + n terms
= n/2 {2 × 1 + ( n - 1) × 2 }
= n²

Tₙ = n²( n + 1)²/4n² = ( n + 1)²/4
S = ∑Tₙ
=1/4 ∑(n + 1)²
= 1/4 (∑n² + 2∑n + ∑1 )

we know,
∑n² = n( n + 1)(2n + 1)/6
∑n = n(n + 1)/2
∑1= n

Sₙ = 1/4 { n(n + 1)(2n + 1)/6 +2n(n + 1)/2 + n }
= 1/4 {n(n + 1)(2n + 1)/6 + n(n + 1) + n}
= n/24 { 2n² + 9n + 13 }

hence, sum of n terms = n/24 { 2n² + 9n + 13 }