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Question Number 9265 by j.masanja06@gmail.com last updated on 26/Nov/16
find the value of        Σ_(k=2) ^5 k^2
$$\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\:\:\:\:\:\:\:\sum_{\mathrm{k}=\mathrm{2}} ^{\mathrm{5}} \mathrm{k}^{\mathrm{2}} \\ $$
Answered by mrW last updated on 27/Nov/16
generally for 1≤m≤n  Σ_(k=1) ^n k^2 =((n(n+1)(2n+1))/6)  Σ_(k=1) ^(m−1) k^2 =(((m−1)m(2m−1))/6)  Σ_(k=m) ^n k^2 =Σ_(k=1) ^n k^2 −Σ_(k=1) ^(m−1) k^2   Σ_(k=m) ^n k^2 =((n(n+1)(2n+1)−(m−1)m(2m−1))/6)    Σ_(k=2) ^5 k^2 =((5(5+1)(2×5+1)−(2−1)2(2×2−1))/6)  =((5×6×11−1×2×3)/6)=((330−6)/6)=54  or  Σ_(k=2) ^5 k^2 =2^2 +3^2 +4^2 +5^2 =4+9+16+25=54
$$\mathrm{generally}\:\mathrm{for}\:\mathrm{1}\leqslant\mathrm{m}\leqslant\mathrm{n} \\ $$$$\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\mathrm{k}^{\mathrm{2}} =\frac{\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)\left(\mathrm{2n}+\mathrm{1}\right)}{\mathrm{6}} \\ $$$$\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{m}−\mathrm{1}} {\sum}}\mathrm{k}^{\mathrm{2}} =\frac{\left(\mathrm{m}−\mathrm{1}\right)\mathrm{m}\left(\mathrm{2m}−\mathrm{1}\right)}{\mathrm{6}} \\ $$$$\underset{\mathrm{k}=\mathrm{m}} {\overset{\mathrm{n}} {\sum}}\mathrm{k}^{\mathrm{2}} =\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\mathrm{k}^{\mathrm{2}} −\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{m}−\mathrm{1}} {\sum}}\mathrm{k}^{\mathrm{2}} \\ $$$$\underset{\mathrm{k}=\mathrm{m}} {\overset{\mathrm{n}} {\sum}}\mathrm{k}^{\mathrm{2}} =\frac{\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)\left(\mathrm{2n}+\mathrm{1}\right)−\left(\mathrm{m}−\mathrm{1}\right)\mathrm{m}\left(\mathrm{2m}−\mathrm{1}\right)}{\mathrm{6}} \\ $$$$ \\ $$$$\underset{\mathrm{k}=\mathrm{2}} {\overset{\mathrm{5}} {\sum}}\mathrm{k}^{\mathrm{2}} =\frac{\mathrm{5}\left(\mathrm{5}+\mathrm{1}\right)\left(\mathrm{2}×\mathrm{5}+\mathrm{1}\right)−\left(\mathrm{2}−\mathrm{1}\right)\mathrm{2}\left(\mathrm{2}×\mathrm{2}−\mathrm{1}\right)}{\mathrm{6}} \\ $$$$=\frac{\mathrm{5}×\mathrm{6}×\mathrm{11}−\mathrm{1}×\mathrm{2}×\mathrm{3}}{\mathrm{6}}=\frac{\mathrm{330}−\mathrm{6}}{\mathrm{6}}=\mathrm{54} \\ $$$$\mathrm{or} \\ $$$$\underset{\mathrm{k}=\mathrm{2}} {\overset{\mathrm{5}} {\sum}}\mathrm{k}^{\mathrm{2}} =\mathrm{2}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} +\mathrm{4}^{\mathrm{2}} +\mathrm{5}^{\mathrm{2}} =\mathrm{4}+\mathrm{9}+\mathrm{16}+\mathrm{25}=\mathrm{54} \\ $$

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