calculate-1-k-1-n-k-2-n-1-k-2-1-i-j-n-ij-

Question Number 73036 by mathmax by abdo last updated on 05/Nov/19

c a l c u l a t e 1) \sum_{k = 1}^{n} k^{2} (n + 1 - k)

2) \sum_{1 ⩽ i ⩽ j ⩽ n} i j

Commented by mathmax by abdo last updated on 06/Nov/19

1) \sum_{k = 1}^{n} k^{2} (n + 1 - k) = (n + 1) \sum_{k = 1}^{n} k^{2} - \sum_{k = 1}^{n} k^{3}

= (n + 1) \times \frac{n (n + 1) (2 n + 1)}{6} - \frac{n^{2} {(n + 1)}^{2}}{4}

= \frac{n (n + 1)}{2} {\frac{(n + 1) (2 n + 1)}{3} - \frac{n (n + 1)}{2}}

= \frac{n (n + 1)}{2} {\frac{2 n^{2} + 3 n + 1}{3} - \frac{n^{2} + n}{2}}

= \frac{n (n + 1)}{12} (4 n^{2} + 6 n + 2 - 3 n^{2} - 3 n)

= \frac{n (n + 1)}{12} (n^{2} + 3 n + 2) = \frac{n (n + 1) (n^{2} + 3 n + 2)}{12}

Commented by mr W last updated on 06/Nov/19

n^{2} + 3 n + 2 = (n + 1) (n + 2)

\Rightarrow w e h a v e t h e s a m e r e s u l t .

Commented by mathmax by abdo last updated on 06/Nov/19

y e s s i r .

Answered by mr W last updated on 06/Nov/19

Σ k^{2} (n + 1 - k)

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

= \frac{n {(n + 1)}^{2} (n + 2)}{12}

Σ Σ i j = Σ i \frac{n (n + 1)}{2} = \frac{n^{2} {(n + 1)}^{2}}{4}

Facebook Tweet Pin