Question and Answers Forum

All Questions      Topic List

Limits Questions

Previous in All Question      Next in All Question      

Previous in Limits      Next in Limits      

Question Number 41454 by Fawomath last updated on 07/Aug/18

$$\mathrm{Evaluate}\underset{k=1}{\sum ^{2n-1}}{(-1)}^{k-1}{k}^3$$

Commented by maxmathsup by imad last updated on 07/Aug/18

$$letS\left(x\right)=\sum _{k=0}^N{(-1)}^k{x}^{k}wehave{S}^{{}^{\prime }}\left(x\right)=\sum _{k=1}^N{(-1)}^{k}k{x}^{k-1}\Rightarrow$$$${xS}^{{}^{\prime }}\left(x\right)=\sum _{k=1}^n{(-1)}^{k}k{x}^k\Rightarrow {\left({xS}^{{}^{\prime }}\left(x\right)\right)}^{{}^{\prime }}=\sum _{k=1}^N{(-1)}^k{k}^2{x}^{k-1}\Rightarrow$$$$S^{{}^{\prime }}\left(x\right)+x{S}^{{}^{″}}\left(x\right)=\sum _{k=1}^N{(-1)}^k{k}^2{x}^{k-1}\Rightarrow$$$${xS}^{{}^{\prime }}\left(x\right)+x^2{S}^{{}^{″}}\left(x\right)=\sum _{k=1}^N{(-1)}^k{k}^2{x}^k\Rightarrow$$$${({xS}^{{}^{\prime }}\left(x\right)+x^2{S}^{{}^{″}}\left(x\right))}^{{}^{\prime }}=\sum _{k=1}^N{(-1)}^k{k}^3{x}^{k-1}\Rightarrow$$$$S^{{}^{\prime }}\left(x\right)+{xS}^{{}^{″}}\left(x\right)+2x{S}^{{}^{″}}\left(x\right)+x^2{S}^{\left(3\right)}\left(x\right)=\sum _{k=1}^N{(-1)}^k{k}^3{x}^{k-1}\Rightarrow$$$${xS}^{{}^{\prime }}\left(x\right)+3x^2{S}^{\left(2\right)}\left(x\right)+x^3{S}^{\left(3\right)}\left(x\right)=\sum _{k=1}^N{(-1)}^k{k}^3{x}^{k}lettakex=1\Rightarrow$$$$\sum _{k=1}^N{(-1)}^k{x}^k=S^{{}^{\prime }}\left(1\right)+3S^{\left(2\right)}\left(1\right)+S^{\left(3\right)}\left(1\right)but$$$$S\left(x\right)=\sum _{k=0}^N{(-x)}^k=\frac{1-{(-x)}^{N+1}}{1+x}=\frac{1-{(-1)}^{N+1}{x}^{N+1}}{1+x}(x\ne -1)$$$$=\frac{1+{(-1)}^N{x}^{N+1}}{1+x}\Rightarrow S^{{}^{\prime }}\left(x\right)=\frac{(N+1){(-1)}^N{x}^N(1+x)-(1+{(-1)}^N{x}^{N+1})}{{(1+x)}^2}$$$$=\frac{(N+1){(-1)}^N{x}^N+(N+1){(-1)}^N{x}^{N+1}-1-{(-1)}^N{x}^{N+1}}{{(1+x)}^2}$$$$=\frac{(N+1){(-1)}^N{x}^N+N{(-1)}^N{x}^{N+1}-1}{{(1+x^{})}^2}\Rightarrow S^{{}^{\prime }}\left(1\right)=\frac{(N+1){(-1)}^N+N{(-1)}^N}{4}$$$$=\frac{(2N+1){(-1)}^N}{4}alsowemustcalculate{S}^{\left(2\right)}\left(1\right)and{S}^{\left(3\right)}\left(1\right)and$$$$takeN=2n-1....$$

Answered by sma3l2996 last updated on 07/Aug/18

$$S_n=\underset{k=1}{\sum ^{2n-1}}{(-1)}^{k-1}{k}^3$$$$=1-2^3+3^3-4^3+...+{(2n-1)}^3$$$$=(1+3^3+5^3+...+{(2n-1)}^3)-(2^3+4^3+6^3+...+{(2n-2)}^3)$$$$=[1+2^3+3^3+4^3+...+{(2n-1)}^3-(2^3+4^3+6^3+...+{(2n-2)}^3)]-2^3(1^3+2^3+3^3+...+{(n-1)}^3)$$$$=\underset{k=1}{\sum ^{2n-1}}{k}^3-2^3\times 2\underset{k=1}{\sum ^{n-1}}{k}^3$$$$={\left[\frac{(2n-1)(2n-1+1)}{2}\right]}^2-2^4{\left[\frac{(n-1)(n-1+1)}{2}\right]}^2$$$$note:\underset{k=1}{\sum ^n}{k}^3={\left[\frac{n(n+1)}{2}\right]}^2$$$$S_n=\frac{{(2n-1)}^2{\left(2n\right)}^2}{2^2}-2^4\frac{{(n-1)}^2{n}^2}{2^2}$$$$=n^2({(2n-1)}^2-2^2{(n-1)}^2)$$$$=n^2(4n^2-4n+1-4n^2+8n-4)$$$$S_n=n^2(4n-3)$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com