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Question Number 38640 by maxmathsup by imad last updated on 27/Jun/18

calculate Σ_(k=1) ^n k^4 interms of n.

$${calculate}\:\sum_{{k}=\mathrm{1}} ^{{n}} {k}^{\mathrm{4}} \:\:{interms}\:{of}\:{n}. \\ $$

Commented by math khazana by abdo last updated on 01/Jul/18

we have (k+1)^5 −k^5 =Σ_(p=0) ^5 C_5 ^p k^p −k^5 1 + C_5 ^1 k +C_5 ^2 k^2 +C_5 ^3 k^3 +C_5 ^4 k^4 +C_5 ^5 k^5 −k^5 =1 +5k +10k^2 + 10k^3 +5 k^4 ⇒ Σ_(k=1) ^n {(k+1)^5 −k^5 } =n +5 Σ_(k=1) ^n k +10 Σ_(k=1) ^n k^2 +10 Σ_(k=1) ^n k^3 +5Σ_(k=1) ^n k^4 (n+1)^5 =n+1 +5 ((n(n+1))/2) +10 ((n(n+1)(2n+1))/6) +10 ((n^2 (n+1)^2 )/4) +5 Σ_(k=1) ^n k^4 after calculus we find Σ_(k=1) ^n k^4 =(n^5 /5) +(n^4 /2) +(n^3 /3) −(n/(30)) .

$${we}\:{have}\:\left({k}+\mathrm{1}\right)^{\mathrm{5}} −{k}^{\mathrm{5}} \\ $$$$=\sum_{{p}=\mathrm{0}} ^{\mathrm{5}} \:\:{C}_{\mathrm{5}} ^{{p}} {k}^{{p}} \:−{k}^{\mathrm{5}} \\ $$$$\mathrm{1}\:\:+\:{C}_{\mathrm{5}} ^{\mathrm{1}} {k}\:\:+{C}_{\mathrm{5}} ^{\mathrm{2}} {k}^{\mathrm{2}} \:\:+{C}_{\mathrm{5}} ^{\mathrm{3}} \:{k}^{\mathrm{3}} \:\:+{C}_{\mathrm{5}} ^{\mathrm{4}} \:{k}^{\mathrm{4}} \:+{C}_{\mathrm{5}} ^{\mathrm{5}} {k}^{\mathrm{5}} \:−{k}^{\mathrm{5}} \\ $$$$=\mathrm{1}\:+\mathrm{5}{k}\:\:+\mathrm{10}{k}^{\mathrm{2}} \:\:+\:\mathrm{10}{k}^{\mathrm{3}} \:\:+\mathrm{5}\:{k}^{\mathrm{4}} \:\Rightarrow \\ $$$$\sum_{{k}=\mathrm{1}} ^{{n}} \left\{\left({k}+\mathrm{1}\right)^{\mathrm{5}} \:−{k}^{\mathrm{5}} \right\} \\ $$$$={n}\:+\mathrm{5}\:\sum_{{k}=\mathrm{1}} ^{{n}} {k}\:+\mathrm{10}\:\sum_{{k}=\mathrm{1}} ^{{n}} \:{k}^{\mathrm{2}} \:+\mathrm{10}\:\sum_{{k}=\mathrm{1}} ^{{n}} \:{k}^{\mathrm{3}} \:+\mathrm{5}\sum_{{k}=\mathrm{1}} ^{{n}} {k}^{\mathrm{4}} \\ $$$$\left({n}+\mathrm{1}\right)^{\mathrm{5}} \:={n}+\mathrm{1}\:+\mathrm{5}\:\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}\:+\mathrm{10}\:\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{6}} \\ $$$$+\mathrm{10}\:\frac{{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{4}}\:+\mathrm{5}\:\sum_{{k}=\mathrm{1}} ^{{n}} \:{k}^{\mathrm{4}} \:{after}\:{calculus}\:{we}\:{find} \\ $$$$\sum_{{k}=\mathrm{1}} ^{{n}} \:{k}^{\mathrm{4}} =\frac{{n}^{\mathrm{5}} }{\mathrm{5}}\:+\frac{{n}^{\mathrm{4}} }{\mathrm{2}}\:+\frac{{n}^{\mathrm{3}} }{\mathrm{3}}\:−\frac{{n}}{\mathrm{30}}\:. \\ $$

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