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Question Number 162074 by naka3546 last updated on 26/Dec/21

$$Findtheexactvalueof$$$$\underset{k=0}{\sum ^{1004}}\left(\underset{k}{\stackrel{2014}{}}\right)\cdot 3^k.$$

Answered by Rasheed.Sindhi last updated on 26/Dec/21

$${(a+b)}^n=\stackrel{n}{\sum _{k=0}}\left(\begin{array}{c}n\\ k\end{array}\right)a^{n-k}{b}^k$$$$Leta=1,b=3\mathrm{\&}n=2014$$$$S\left(2015\right)={(1+3)}^{2014}=\underset{k=0}{\sum ^{2014}}\left(\begin{array}{c}2014\\ k\end{array}\right).3^k=4^{2014}....\left(i\right)$$$${We}^{\prime }vetodeterminefirst1005terms\left(sum\right)$$$$i-e\underset{k=0}{\sum ^{1004}}\left(\underset{k}{\stackrel{2014}{}}\right)\cdot 3^k$$$$S\left(1\mathrm{to}2015\right)=\underset{{\mathrm{S}}_1}{\underbrace{\{{\mathrm{t}}_1+{\mathrm{t}}_2+...{\mathrm{t}}_{1007}+\frac{{\mathrm{t}}_{1008}}{2}\}}}$$$$+\underset{S_2}{\underbrace{\{\frac{{\mathrm{t}}_{1008}}{2}+{\mathrm{t}}_{1009}+{\mathrm{t}}_{1010}+...+{\mathrm{t}}_{2015}\}}}$$$$\because Here{S}_1=S_2\therefore S_1=S/2=4^{2014}/2$$$$\Rightarrow S_1=4^{2013}$$$$\therefore S\left(1\mathrm{to}1005\right)=S_1-({\mathrm{t}}_{1006}+{\mathrm{t}}_{1007}+\frac{{\mathrm{t}}_{1008}}{2})$$$$=4^{2013}-(\left(\begin{array}{c}2014\\ 1005\end{array}\right).3^{1005}+\left(\begin{array}{c}2014\\ 1006\end{array}\right).3^{1006}+\frac{1}{2}\left(\begin{array}{c}2014\\ 1007\end{array}\right).3^{1007})$$$$=4^{2013}-3^{1005}(\left(\begin{array}{c}2014\\ 1005\end{array}\right)+3\left(\begin{array}{c}2014\\ 1006\end{array}\right)+\frac{9}{2}\left(\begin{array}{c}2014\\ 1007\end{array}\right))$$

Commented by naka3546 last updated on 26/Dec/21

$$Thankyou,sir.$$

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