Question

Derive a formula for Summation n i=1 i³ by using a telescoping sum with terms f(i)=i²

Answer 1

i⁴ - (i-1)⁴ = i⁴ - (i⁴ - 4i³ + 6i² - 4i + 1)

=> i⁴ - (i-1)⁴ = i⁴ - i⁴ + 4i³ - 6i² + 4i - 1

=> i⁴ - (i-1)⁴ = 4i³ - 6i² + 4i - 1

Taking sum over i = 1 to n

$\sum_{i=1}^{n}$ {i⁴ - (i-1)⁴ }= $\sum_{i=1}^{n}${4i³- 6i²+4i - 1}

=> n⁴ - 0⁴ = $\sum_{i=1}^{n}4i³-\sum_{i=1}^{n} 6i² +\sum_{i=1}^{n} 4i - \sum_{i=1}^{n} 1$

=> n⁴ = $4\sum_{i=1}^{n}i³-6\sum_{i=1}^{n} i² +4\sum_{i=1}^{n} i - \sum_{i=1}^{n} 1$

=> n⁴ = $4\sum_{i=1}^{n}i³-$ 6.$\frac{n(n+1)(2n+1)}{6}$ + $4.\frac{n(n+1)}{2}-n$

=> n⁴ = $4\sum_{i=1}^{n}i³$ - n(n+1)(2n+1) + 2n(n+1) - n

=> $4\sum_{i=1}^{n}i³$ = n⁴ + n(n+1)(2n+1) - 2n(n+1) + n

=> $4\sum_{i=1}^{n}i³$ = n⁴ + 2n³ + 3n² + n - 2n² - 2n + n

=> $4\sum_{i=1}^{n}i³$ = n⁴ + 2n³ + n²

=> $4\sum_{i=1}^{n}i³$ = (n² + n)²

=> $4\sum_{i=1}^{n}i³$ = {n(n+1)}²

=> $\sum_{i=1}^{n}i³$ = $\frac{ {n²(n+1)}²}{4}$

=> $\sum_{i=1}^{n}i³ =$ $[\frac{n(n+1)}{2}]²$