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Question Number 49937 by turbo msup by abdo last updated on 12/Dec/18

$$let{u}_n=\frac{{(-1)}^n}{2^n}+3n+1$$$$find\sum _{n=0}^{49}{u}_n$$

Commented by Abdo msup. last updated on 13/Dec/18

$$wehave{u}_n=v_n+w_{n}with{v}_n={(-\frac{1}{2})}^{n}geometricwithq=-\frac{1}{2}$$$$and{w}_n=3n+1srithmeticwithr=3$$$$\sum _{n=0}^{49}{u}_n=\sum _{n=0}^{49}{u}_n+\sum _{n=0}^{49}{w}_n$$$$=u_0\frac{1-q^{50}}{1-q}+\frac{50}{2}(w_0+w_n)$$$$=\frac{1-{(-\frac{1}{2})}^{50}}{1+\frac{1}{2}}+25(1+3n+1)$$$$=\frac{2}{3}(1-\frac{1}{2^{50}})+25(3n+2).$$

Commented by Abdo msup. last updated on 13/Dec/18

$$n=49\Rightarrow \sum _{n=0}^{49}{u}_n=\frac{2}{3}(1-\frac{1}{2^{50}})+25(3.49+2)$$$$=\frac{2}{3}(1-\frac{1}{2^{50}})+25.147+2=....$$

Answered by afachri last updated on 12/Dec/18

$$\underset{n=0}{\sum ^{49}}\frac{{(-1)}^n}{2^n}+3n+1=\underset{n=1}{\sum ^{50}}{\left(2^{-1}\right)}^{n-1}+3(n-1)+1$$$$=\underset{n=1}{\sum ^{50}}[{}_{}^{}{\left(2\right)}^{1-n}+3n-2{}_{}^{}]$$$$=\underset{n=1}{\sum ^{50}}(\frac{2}{2^n}+3n-2)$$$$\mathrm{Split}\mathrm{the}\mathrm{eq}\mathrm{in}\mathrm{three}\mathrm{terms}:$$$$=\underset{n=1}{\sum ^{50}}\left(\frac{2}{2^n}\right)+3\underset{n=1}{\sum ^{50}}\left(n\right)-\underset{n=1}{\sum ^{50}}\left(2\right)$$$$\mathbf{note}:\frac{2}{2^n}\mathbf{is}\mathbf{geometric}\mathbf{series}\mathbf{with}\mathit{r}=-\frac{1}{2}$$$$\mathbf{so}\mathbf{the}\mathbf{sigma}\mathbf{will}\mathbf{be}\mathbf{its}\mathbf{sum}.S_{50}=\frac{a(1-r^n)}{1-r}$$$$=\left(\frac{1(1-{(-\frac{1}{2})}^{50})}{1-(-\frac{1}{2})}\right)+3\left(\frac{50\left(51\right)}{2}\right)-2\left(50\right)$$$$=\frac{2}{3}\left(\frac{2^{50}-1}{2^{50}}\right)+3825-100$$$$\mathrm{The}\mathrm{result}\mathrm{for}\frac{2^{50}-1}{2^{50}}\mathrm{is}\approx 1$$$$=\frac{2\left(1\right)}{3}+3725$$$$=3725\frac{2}{3}\left[\mathrm{Done}\right]$$$$$$$$$$$$$$$$$$

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