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Question Number 113015 by mohssinee last updated on 10/Sep/20

Answered by Olaf last updated on 10/Sep/20

Σ_(k=0) ^n (2k^2 +3k+5) = 2Σ_(k=0) ^n k^2 +3Σ_(k=0) ^n k+Σ_(k=0) ^n 5 = 2((n(n+1)(2n+1))/6)+3((n(n+1))/2)+5(n+1) = (n+1)[((n(2n+1))/3)+(3/2)n+5] = (n+1)[(2/3)n^2 +(1/3)n+(3/2)n+5] = (n+1)[(2/3)n^2 +((11)/6)n+5] = (1/6)(n+1)(4n^2 +11n+30)

$$\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\left(\mathrm{2}{k}^{\mathrm{2}} +\mathrm{3}{k}+\mathrm{5}\right)\:= \\ $$$$\mathrm{2}\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}{k}^{\mathrm{2}} +\mathrm{3}\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}{k}+\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\mathrm{5}\:= \\ $$$$\mathrm{2}\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{6}}+\mathrm{3}\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}+\mathrm{5}\left({n}+\mathrm{1}\right) \\ $$$$=\:\left({n}+\mathrm{1}\right)\left[\frac{{n}\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{3}}+\frac{\mathrm{3}}{\mathrm{2}}{n}+\mathrm{5}\right]\:= \\ $$$$\left({n}+\mathrm{1}\right)\left[\frac{\mathrm{2}}{\mathrm{3}}{n}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{3}}{n}+\frac{\mathrm{3}}{\mathrm{2}}{n}+\mathrm{5}\right]\:= \\ $$$$\left({n}+\mathrm{1}\right)\left[\frac{\mathrm{2}}{\mathrm{3}}{n}^{\mathrm{2}} +\frac{\mathrm{11}}{\mathrm{6}}{n}+\mathrm{5}\right]\:= \\ $$$$\frac{\mathrm{1}}{\mathrm{6}}\left({n}+\mathrm{1}\right)\left(\mathrm{4}{n}^{\mathrm{2}} +\mathrm{11}{n}+\mathrm{30}\right) \\ $$

Answered by mathmax by abdo last updated on 10/Sep/20

Σ_(k=0) ^n (2k^2 +3k +5) =2Σ_(k=0) ^n k^2 +3Σ_(k=0) ^n k +5Σ_(k=0) ^n (1) =2×((n(n+1)(2n+1))/6) +3.((n(n+1))/2) +5(n+1) =(n+1){((n(2n+1))/3) +((3n)/2) +5} =(n+1)( ((2n(2n+1)+9n +30)/6)) =((n+1)/6)(4n^2 +2n+9n +30) =(((n+1)(4n^2 +11n +30))/6)

$$\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{n}} \left(\mathrm{2k}^{\mathrm{2}} \:+\mathrm{3k}\:+\mathrm{5}\right)\:=\mathrm{2}\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{n}} \:\mathrm{k}^{\mathrm{2}} \:+\mathrm{3}\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{n}} \:\mathrm{k}\:+\mathrm{5}\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{n}} \left(\mathrm{1}\right) \\ $$$$=\mathrm{2}×\frac{\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)\left(\mathrm{2n}+\mathrm{1}\right)}{\mathrm{6}}\:+\mathrm{3}.\frac{\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)}{\mathrm{2}}\:+\mathrm{5}\left(\mathrm{n}+\mathrm{1}\right) \\ $$$$=\left(\mathrm{n}+\mathrm{1}\right)\left\{\frac{\mathrm{n}\left(\mathrm{2n}+\mathrm{1}\right)}{\mathrm{3}}\:+\frac{\mathrm{3n}}{\mathrm{2}}\:+\mathrm{5}\right\} \\ $$$$=\left(\mathrm{n}+\mathrm{1}\right)\left(\:\frac{\mathrm{2n}\left(\mathrm{2n}+\mathrm{1}\right)+\mathrm{9n}\:+\mathrm{30}}{\mathrm{6}}\right)\:=\frac{\mathrm{n}+\mathrm{1}}{\mathrm{6}}\left(\mathrm{4n}^{\mathrm{2}} \:+\mathrm{2n}+\mathrm{9n}\:+\mathrm{30}\right) \\ $$$$=\frac{\left(\mathrm{n}+\mathrm{1}\right)\left(\mathrm{4n}^{\mathrm{2}} \:+\mathrm{11n}\:+\mathrm{30}\right)}{\mathrm{6}} \\ $$

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