Question and Answers Forum

All Questions      Topic List

Differentiation Questions

Previous in All Question      Next in All Question      

Previous in Differentiation      Next in Differentiation      

Question Number 36099 by rahul 19 last updated on 28/May/18

Commented by rahul 19 last updated on 01/Jun/18

$$\mathrm{Help}\mathrm{in}\mathrm{Q}.1.$$

Commented by rahul 19 last updated on 31/Aug/18

$$\mathrm{Thank}\mathrm{you}\mathrm{prof}\mathrm{Abdo}:)$$

Commented by maxmathsup by imad last updated on 31/Aug/18

$$2)wehavef(x+1)={(-1)}^{x+1}x-2f\left(x\right)xfromNf\left(1\right)=f\left(1986\right)$$$$letfindS=f\left(1\right)+f\left(2\right)+....f\left(1985\right)$$$$wehave\sum _{k=1}^{1985}f(k+1)=\sum _{k=1}^{1985}k{(-1)}^{k+1}-2\sum _{k=1}^{1985}f\left(k\right)\Rightarrow$$$$\sum _{k=2}^{1986}f\left(k\right)=\sum _{k=1}^{1985}k{(-1)}^{k+1}-2S\Rightarrow \sum _{k=1}^{1985}f\left(k\right)+f\left(1986\right)-f\left(1\right)$$$$=-2S-\sum _{k=1}^{1985}k{(-1)}^k\Rightarrow 3S=-\sum _{k=1}^{1985}k{(-1)}^{k}forx\ne 1$$$$letp\left(x\right)=\sum _{k=0}^n{x}^k\Rightarrow p^{{}^{\prime }}\left(x\right)=\sum _{k=1}^{n}k{x}^{k-1}\Rightarrow x{p}^{{}^{\prime }}\left(x\right)=\sum _{k=1}^{n}k{x}^{k}but$$$$p\left(x\right)=\frac{x^{n+1}-1}{x-1}\Rightarrow p^{{}^{\prime }}\left(x\right)=\frac{{nx}^{n+1}-(n+1)x^n+1}{{(x-1)}^2}\Rightarrow$$$$\sum _{k=1}^{n}k{x}^k=\frac{x}{{(x-1)}^2}\{{nx}^{n+1}-(n+1)x^n+1\}\Rightarrow$$$$\sum _{k=1}^{n}k{(-1)}^k=\frac{-1}{4}\{n{(-1)}^{n+1}-(n+1){(-1)}^n+1\}$$$$=-\frac{1}{4}\{1-(2n+1){(-1)}^n\}=\frac{(2n+1){(-1)}^n-1}{4}\Rightarrow$$$$\sum _{k=1}^{1985}k{(-1)}^k=\frac{-(2\times 1985+1)-1}{4}=\frac{-3972}{4}=-993\Rightarrow$$$$3S=993\Rightarrow S=331.$$$$$$$$$$

Commented by maxmathsup by imad last updated on 31/Aug/18

$$nevermindsir.$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com