Question

1+1/2^3+1/3^3+1/4^3+----

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 }

Answer 2

Answer:

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 )

Step-by-step explanation: