Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 103560 by abony1303 last updated on 15/Jul/20

$$\mathrm{S}=\underset{\mathrm{k}=1}{\sum ^{17}}\mathrm{k}\cdot 2^{\mathrm{k}}=?$$

Commented by abony1303 last updated on 15/Jul/20

$$\mathrm{pls}\mathrm{help}$$

Answered by mr W last updated on 15/Jul/20

$$S_n=\underset{k=1}{\sum ^n}k{2}^k$$$$2S_n=\underset{k=1}{\sum ^n}k{2}^{k+1}$$$$2S_n=\underset{k=1}{\sum ^n}(k+1)2^{k+1}-\underset{k=1}{\sum ^n}{2}^{k+1}$$$$2S_n=\underset{k=2}{\sum ^{n+1}}k{2}^k-2\underset{k=1}{\sum ^n}{2}^k$$$$2S_n=\underset{k=1}{\sum ^n}k{2}^k+(n+1)2^{n+1}-2-2\times \frac{2(2^n-1)}{2-1}$$$$2S_n=S_n+(n-1)2^{n+1}+2$$$$\Rightarrow S_n=(n-1)2^{n+1}+2$$

Answered by Dwaipayan Shikari last updated on 15/Jul/20

$$S_n=1.2+2.2^2+3.2^3+...........+n.2^n$$$$$$$$2S_n=1.2^2+2.2^3+.....+(n-1)2^n+n{2}^{n+1}$$$$............Subtracting$$$$-S_n=1.2+1.2^2+1.2^3+....+2^n-n{2}^{n+1}$$$$S_n=n{2}^{n+1}-(1.2+\frac{2^n-1}{1})$$$$S_n={(n-1)}^{2n+1}+2$$$$S_{17}=16.2^{18}+2(⊛◼★\mathrm{DS})$$

Answered by OlafThorendsen last updated on 15/Jul/20

$$f\left(x\right)=\underset{k=0}{\sum ^n}{x}^k=\frac{1-x^{n+1}}{1-x}$$$$f^{\prime }\left(x\right)=\underset{k=1}{\sum ^n}{kx}^{k-1}$$$$\mathrm{and}{f}^{\prime }\left(x\right)=\frac{-(n+1)x^n(1-x)+(1-x^{n+1})}{{(1-x)}^2}$$$${xf}^{\prime }\left(x\right)=\underset{k=1}{\sum ^n}{kx}^k=x\frac{1-(n+1)x^n+{nx}^{n+1}}{{(1-x)}^2}$$$$\Rightarrow \underset{k=1}{\sum ^{17}}k{2}^k=2\frac{1-(17+1)2^{17}+17.2^{17+1}}{{(1-2)}^2}$$$$\Rightarrow \underset{k=1}{\sum ^{17}}k{2}^k=2(17.2^{18}-18.2^{17}+1)$$$$\Rightarrow \underset{k=1}{\sum ^{17}}k{2}^k=4.194.306$$$$$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com