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

calculate 1) Σ_(k=1) ^n k^2 (n+1−k) 2)Σ_(1≤i≤j≤n) ij

$$\left.{calculate}\:\mathrm{1}\right)\:\sum_{{k}=\mathrm{1}} ^{{n}} \:{k}^{\mathrm{2}} \left({n}+\mathrm{1}−{k}\right) \\ $$$$\left.\mathrm{2}\right)\sum_{\mathrm{1}\leqslant{i}\leqslant{j}\leqslant{n}} \:{ij} \\ $$

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))

$$\left.\mathrm{1}\right)\:\sum_{{k}=\mathrm{1}} ^{{n}} {k}^{\mathrm{2}} \left({n}+\mathrm{1}−{k}\right)\:=\left({n}+\mathrm{1}\right)\sum_{{k}=\mathrm{1}} ^{{n}} {k}^{\mathrm{2}} \:−\sum_{{k}=\mathrm{1}} ^{{n}} \:{k}^{\mathrm{3}} \\ $$$$=\left({n}+\mathrm{1}\right)×\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{6}}−\frac{{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{4}} \\ $$$$=\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}\left\{\frac{\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{3}}−\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}\right\} \\ $$$$=\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}\left\{\frac{\mathrm{2}{n}^{\mathrm{2}} +\mathrm{3}{n}+\mathrm{1}}{\mathrm{3}}−\frac{{n}^{\mathrm{2}} \:+{n}}{\mathrm{2}}\right\} \\ $$$$=\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{12}}\left(\mathrm{4}{n}^{\mathrm{2}} \:+\mathrm{6}{n}+\mathrm{2}−\mathrm{3}{n}^{\mathrm{2}} −\mathrm{3}{n}\right) \\ $$$$=\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{12}}\left({n}^{\mathrm{2}} +\mathrm{3}{n}+\mathrm{2}\right)\:=\frac{{n}\left({n}+\mathrm{1}\right)\left({n}^{\mathrm{2}} \:+\mathrm{3}{n}+\mathrm{2}\right)}{\mathrm{12}} \\ $$

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

n^2 +3n+2=(n+1)(n+2) ⇒we have the same result.

$${n}^{\mathrm{2}} +\mathrm{3}{n}+\mathrm{2}=\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right) \\ $$$$\Rightarrow{we}\:{have}\:{the}\:{same}\:{result}. \\ $$

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

$${yes}\:{sir}. \\ $$

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

Σk^2 (n+1−k) =(n+1)((n(n+1)(2n+1))/6)−((n^2 (n+1)^2 )/4) =((n(n+1)^2 (n+2))/(12)) ΣΣij=Σi((n(n+1))/2)=((n^2 (n+1)^2 )/4)

$$\Sigma{k}^{\mathrm{2}} \left({n}+\mathrm{1}−{k}\right) \\ $$$$=\left({n}+\mathrm{1}\right)\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{6}}−\frac{{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{4}} \\ $$$$=\frac{{n}\left({n}+\mathrm{1}\right)^{\mathrm{2}} \left({n}+\mathrm{2}\right)}{\mathrm{12}} \\ $$$$ \\ $$$$\Sigma\Sigma{ij}=\Sigma{i}\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}=\frac{{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{4}} \\ $$

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