calculate 1) Σ_(k=1) ^n k^2 (n+1−k) 2)Σ


Question and Answers Forum

All Questions Topic List
Algebra Questions
Previous in All Question Next in All Question
Previous in Algebra Next in Algebra
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
$1) Σ_(k=1) ^n k^2 (n+1−k) =(n+1)Σ_(k=1) ^n k^2 −Σ_(k=1) ^n k^3 =(n+1)×((n(n+1)(2n+1))/6)−((n^2 (n+1)^2 )/4) =((n(n+1))/2){(((n+1)(2n+1))/3)−((n(n+1))/2)} =((n(n+1))/2){((2n^2 +3n+1)/3)−((n^2 +n)/2)} =((n(n+1))/(12))(4n^2 +6n+2−3n^2 −3n) =((n(n+1))/(12))(n^2 +3n+2) =((n(n+1)(n^2 +3n+2))/(12))$
$1) \sum_{k = 1}^{n} k^{2} (n + 1 - k) = (n + 1) \sum_{k = 1}^{n} k^{2} - \sum_{k = 1}^{n} k^{3}$ $= (n + 1) \times \frac{n (n + 1) (2 n + 1)}{6} - \frac{n^{2} {(n + 1)}^{2}}{4}$ $= \frac{n (n + 1)}{2} {\frac{(n + 1) (2 n + 1)}{3} - \frac{n (n + 1)}{2}}$ $= \frac{n (n + 1)}{2} {\frac{2 n^{2} + 3 n + 1}{3} - \frac{n^{2} + n}{2}}$ $= \frac{n (n + 1)}{12} (4 n^{2} + 6 n + 2 - 3 n^{2} - 3 n)$ $= \frac{n (n + 1)}{12} (n^{2} + 3 n + 2) = \frac{n (n + 1) (n^{2} + 3 n + 2)}{12}$
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

	$Σ k^{2} (n + 1 - k)$ $= (n + 1) \frac{n (n + 1) (2 n + 1)}{6} - \frac{n^{2} {(n + 1)}^{2}}{4}$ $= \frac{n {(n + 1)}^{2} (n + 2)}{12}$ $Σ Σ i j = Σ i \frac{n (n + 1)}{2} = \frac{n^{2} {(n + 1)}^{2}}{4}$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum