Question

Derive a formula for
n
∑
i= i² a telescoping sum with terms f(I) = i³


n
∑ (3i-1)²
i=1


n
∑ (2i + 1)²
i=1

Answer 1

$\sum_{i=1}^{n} i$

= 1+2+3+......+n

= n(n+1)/2 by formula of sum of AP

(i+1)³ - i³ = 3i²+3i+1

$\sum_{i=1}^{n} (i+1)^3 -i^3 = \\\sum_{i=1}^{n} 3i²+\sum_{i=1}^{n} 3i + \sum_{i=1}^{n} 1$

=>

$\sum_{i=1}^{n} (i+1)^3 -i^3 =\\ 3\sum_{i=1}^{n} i²+3\sum_{i=1}^{n} i + \sum_{i=1}^{n} 1$

=>

2³-1³+3³-2³+4³-3³....(n+1)³-n³ =

$3\sum_{i=1}^{n} i²+3\sum_{i=1}^{n} i + \sum_{i=1}^{n} 1$

=>

(i+1)³ - i³ = $3\sum_{i=1}^{n} i²+3\sum_{i=1}^{n} i + \sum_{i=1}^{n} 1$

So

$3\sum_{i=1}^{n} i² \\= (n+1)³ - 1 - 3\sum_{i=1}^{n} i - \sum_{i=1}^{n} 1$

= n³ + 3n² + 3n + 1 - 1 - 3n(n+1)/2 -n

= $\frac{2n³+3n²+n}{2}$

= $\frac{n(n+1)(2n+1)}{2}$

So $\sum_{I=1}^{n} i² =$ $\frac{n(n+1)(2n+1)}{6}$

Now

(i+1)⁴ - i⁴= 4i³+6i²+4i+1

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

=> 2⁴-1⁴+3⁴-2⁴.....(n+1)⁴-n⁴

= 4$\sum_{1}^{n} i³$ + 6$\frac{n(n+1)(2n+1)}{6}$ + 4 $\frac{n(n+1)}{2}$ +n

=> 4$\sum_{1}^{n} i³$

= (n+1)⁴-1-n(n+1)(2n+1) - 2n² - 2n - n

= (n+1)⁴ - (n+1)(2n²+n+2n+1)

= (n+1)⁴ - (n+1)(2n+1)(n+1)

= (n+1)²(n²+2n+1-2n-1)

= n²(n+1)²

So

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

$\sum_{i=1}^{n} (3i-1)²$ $=\sum_{i=1}^{n} (9i²-6i+1)$

$=9\sum_{i=1}^{n} i²-6\sum_{i=1}^{n} i + \sum_{i=1}^{n} 1$

= 9 $\frac{n(n+1)(2n+1)}{6}$ - 6 $\frac{n(n+1)}{2}$ + n

= $\frac{n(6n²+3n-1)}{2}$

$\sum_{i=1}^{n} (2i+1)²$

= $\sum_{i=1}^{n} (4i²+4i+1)$

= $4\sum_{i=1}^{n} i²+4\sum_{i=1}^{n} i + \sum_{i=1}^{n} 1$

= 4$\frac{n(n+1)(2n+1)}{6}$ + 4$\frac{n(n+1)}{2}$ + n

= $\frac{n(4n²+12n+11)}{3}$