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Question Number 188786 by Rupesh123 last updated on 07/Mar/23

Commented by Rupesh123 last updated on 07/Mar/23

Prove that:

Answered by mr W last updated on 07/Mar/23

$${(1+x)}^n{(1+x)}^n={(1+x)}^{2n}$$$$\left[\underset{k=0}{\sum ^n}{x}^k\left(\begin{array}{c}n\\ k\end{array}\right)\right]\left[\underset{r=0}{\sum ^n}{x}^r\left(\begin{array}{c}n\\ r\end{array}\right)\right]={(1+x)}^{2n}$$$$\left[\underset{k=0}{\sum ^n}{kx}^{k-1}\left(\begin{array}{c}n\\ k\end{array}\right)\right]\left[\underset{r=0}{\sum ^n}{x}^r\left(\begin{array}{c}n\\ r\end{array}\right)\right]+\left[\underset{k=0}{\sum ^n}{x}^k\left(\begin{array}{c}n\\ k\end{array}\right)\right]\left[\underset{r=0}{\sum ^n}{rx}^{r-1}\left(\begin{array}{c}n\\ r\end{array}\right)\right]=2n{(1+x)}^{2n-1}$$$$\left[\underset{k=0}{\sum ^n}{kx}^{k-1}\left(\begin{array}{c}n\\ k\end{array}\right)\right]\left[\underset{r=0}{\sum ^n}{x}^r\left(\begin{array}{c}n\\ r\end{array}\right)\right]=n{(1+x)}^{2n-1}$$$$\left[\underset{k=0}{\sum ^n}{kx}^{k-1}\left(\begin{array}{c}n\\ k\end{array}\right)\right]\left[\underset{r=0}{\sum ^n}{x}^r\left(\begin{array}{c}n\\ r\end{array}\right)\right]=n\underset{k=0}{\sum ^{2n-1}}{x}^k\left(\begin{array}{c}2n-1\\ k\end{array}\right)$$$$coef.of{x}^{n-1}:$$$$fromRHS=n\left(\begin{array}{c}2n-1\\ n-1\end{array}\right)$$$$fromLHS=\underset{k=0}{\sum ^n}k\left(\begin{array}{c}n\\ k\end{array}\right)\left(\begin{array}{c}n\\ n-k\end{array}\right)=\underset{k=1}{\sum ^n}k{\left(\begin{array}{c}n\\ k\end{array}\right)}^2$$$$\Rightarrow \underset{k=1}{\sum ^n}k{\left(\begin{array}{c}n\\ k\end{array}\right)}^2=n\left(\begin{array}{c}2n-1\\ n-1\end{array}\right)$$

Commented by Rupesh123 last updated on 07/Mar/23

Nice solution, sir!

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