Question

please answer this Q

Answer 1

Answer: 2n^4 - n² or n² (2n² - 1)

Solution:

The nth term of the given series

1³ + 5³ + 7³ +…….. ……….

……………………………………………………………………………………………(1)

is, nth term = (2n - 1)³ = 8n³ - 1 - 3.2n.1 (2n-1) = 8n³ - 12n² +6n - 1

Let the sum up to n terms of the series (1) be denoted by S; then

S = 8 ∑n³ - 12 ∑n² +6 ∑n - ∑1

where

∑n³ = 1³ + 2³ + 3³ +…….. ……….+n³

∑n² = 1² + 2² + 3² +…………….+n²

∑n = 1 + 2 + 3 + 4 +…………….+n

and the summation runs from 1 to n . But we know that

∑n³ = n² (n+1)²/4 , ∑n² = n(n+1) (2n+1)/6 and ∑n = n(n+1)/2

Substituting the above formulae and noting that ∑1 = 1+1+1+…… n times = n,

S = 8 . n² (n+1)²/4 - 12 . n . (n+1) . (2n+1)/6 + 6 . n(n+1)/2 - n

= 2 n² (n²+2n+1) - 2 (n²+n) . (2n+1) + 3 (n²+n) - n

= 2n^4 + 4n³ + 2n² - 2(2n³ + n² + 2n² + n) + 3n² + 3n - n

= 2n^4 + 4n³ + 2n² - 4n³ - 2n² - 4n² - 2n + 3n² + 3n - n Cancelling out equal terms,

= 2n^4 - 4n² - 2n + 3n² + 3n - n Simplifying further

S = 2n^4 - n² or n² (2n² - 1) (Proved)