Question

Prove that:
1/(n+1) + 1/2*(n+1)² + 1/3*(n+1)³ +....... = 1/n -1/2*n² + 1/3*n³-.......
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Answer 1

Step-by-step explanation:

Though not expressly stated in the question, we will assume here that n stands for the first n natural numbers.

∴ Summation of (n-1)³

= ∑(n-1)³

= ∑(n³ - 3n².1 +3.n.1² - 1³) [On using the formula (a-b)³=a³ - 3a²b + 3ab² - b³]

=∑(n³ - 3n² + 3n - 1)

=∑n³ - 3∑n² + 3∑n - ∑1……………………………………………….…….……(1)

We have

∑n³ = sum of the cubes of the first n natural numbers ={n(n+1)/2}²

∑n² = sum of the squares of the first n natural numbers=n(n+1)(2n+1)/6