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Question Number 9279 by j.masanja06@gmail.com last updated on 28/Nov/16

evalute the value of Σ_(m=2 ) ^5 m^4

$$\mathrm{evalute}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\: \\ $$$$\overset{\mathrm{5}} {\sum}_{\mathrm{m}=\mathrm{2}\:} \mathrm{m}^{\mathrm{4}} \\ $$

Answered by mrW last updated on 28/Nov/16

please refer to Q9232 and Q9265.

$$\mathrm{please}\:\mathrm{refer}\:\mathrm{to}\:\mathrm{Q9232}\:\mathrm{and}\:\mathrm{Q9265}. \\ $$

Answered by geovane10math last updated on 28/Nov/16

2^4 + 3^4 + 4^4 + 5^4 + 6^4 = 2^4 = 16 3^4 = 81 4^4 = 256 5^4 = 625 6^4 = 1296 16 + 81 + 256 + 625 + 1296 = = 97 + 881 + 1296 = 1296 + 978 = 2274

$$\mathrm{2}^{\mathrm{4}} \:+\:\mathrm{3}^{\mathrm{4}} \:+\:\mathrm{4}^{\mathrm{4}} \:+\:\mathrm{5}^{\mathrm{4}} \:+\:\mathrm{6}^{\mathrm{4}} \:=\: \\ $$$$\mathrm{2}^{\mathrm{4}} \:=\:\mathrm{16} \\ $$$$\mathrm{3}^{\mathrm{4}} \:=\:\mathrm{81} \\ $$$$\mathrm{4}^{\mathrm{4}} \:=\:\mathrm{256} \\ $$$$\mathrm{5}^{\mathrm{4}} \:=\:\mathrm{625} \\ $$$$\mathrm{6}^{\mathrm{4}} \:=\:\mathrm{1296} \\ $$$$\mathrm{16}\:+\:\mathrm{81}\:+\:\mathrm{256}\:+\:\mathrm{625}\:+\:\mathrm{1296}\:=\: \\ $$$$=\:\mathrm{97}\:+\:\mathrm{881}\:+\:\mathrm{1296}\:=\:\mathrm{1296}\:+\:\mathrm{978}\:=\: \\ $$$$\mathrm{2274} \\ $$

Answered by mrW last updated on 28/Nov/16

generally for 1≤m≤n Σ_(k=1) ^n k^4 =((n(n+1)(2n+1)(3n^2 +3n−1))/(30)) Σ_(k=1) ^(m−1) k^4 =(((m−1)m(2m−1)(3m^2 −3m−1))/(30)) Σ_(k=m) ^n k^4 =Σ_(k=1) ^n k^4 −Σ_(k=1) ^(m−1) k^4 Σ_(k=m) ^n k^4 =((n(n+1)(2n+1)(3n^2 +3n−1)−(m−1)m(2m−1)(3m^2 −3m−1))/(30)) Σ_(k=2) ^5 k^4 =((5×6×11×(3×5^2 +3×5−1)−1×2×3×(3×2^2 −3×2−1))/(30)) =((5×6×11×89−6×5)/(30))=978 or 2^4 +3^4 +4^4 +5^4 =16+81+256+625=978

$$\mathrm{generally}\:\mathrm{for}\:\mathrm{1}\leqslant\mathrm{m}\leqslant\mathrm{n} \\ $$$$\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\mathrm{k}^{\mathrm{4}} =\frac{\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)\left(\mathrm{2n}+\mathrm{1}\right)\left(\mathrm{3n}^{\mathrm{2}} +\mathrm{3n}−\mathrm{1}\right)}{\mathrm{30}} \\ $$$$\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{m}−\mathrm{1}} {\sum}}\mathrm{k}^{\mathrm{4}} =\frac{\left(\mathrm{m}−\mathrm{1}\right)\mathrm{m}\left(\mathrm{2m}−\mathrm{1}\right)\left(\mathrm{3m}^{\mathrm{2}} −\mathrm{3m}−\mathrm{1}\right)}{\mathrm{30}} \\ $$$$\underset{\mathrm{k}=\mathrm{m}} {\overset{\mathrm{n}} {\sum}}\mathrm{k}^{\mathrm{4}} =\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\mathrm{k}^{\mathrm{4}} −\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{m}−\mathrm{1}} {\sum}}\mathrm{k}^{\mathrm{4}} \\ $$$$\underset{\mathrm{k}=\mathrm{m}} {\overset{\mathrm{n}} {\sum}}\mathrm{k}^{\mathrm{4}} =\frac{\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)\left(\mathrm{2n}+\mathrm{1}\right)\left(\mathrm{3n}^{\mathrm{2}} +\mathrm{3n}−\mathrm{1}\right)−\left(\mathrm{m}−\mathrm{1}\right)\mathrm{m}\left(\mathrm{2m}−\mathrm{1}\right)\left(\mathrm{3m}^{\mathrm{2}} −\mathrm{3m}−\mathrm{1}\right)}{\mathrm{30}} \\ $$$$\underset{\mathrm{k}=\mathrm{2}} {\overset{\mathrm{5}} {\sum}}\mathrm{k}^{\mathrm{4}} =\frac{\mathrm{5}×\mathrm{6}×\mathrm{11}×\left(\mathrm{3}×\mathrm{5}^{\mathrm{2}} +\mathrm{3}×\mathrm{5}−\mathrm{1}\right)−\mathrm{1}×\mathrm{2}×\mathrm{3}×\left(\mathrm{3}×\mathrm{2}^{\mathrm{2}} −\mathrm{3}×\mathrm{2}−\mathrm{1}\right)}{\mathrm{30}} \\ $$$$=\frac{\mathrm{5}×\mathrm{6}×\mathrm{11}×\mathrm{89}−\mathrm{6}×\mathrm{5}}{\mathrm{30}}=\mathrm{978} \\ $$$$\mathrm{or} \\ $$$$\mathrm{2}^{\mathrm{4}} +\mathrm{3}^{\mathrm{4}} +\mathrm{4}^{\mathrm{4}} +\mathrm{5}^{\mathrm{4}} =\mathrm{16}+\mathrm{81}+\mathrm{256}+\mathrm{625}=\mathrm{978} \\ $$

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