If total number of runs scored in n matches isn ( (n + 1 / 4) ) n ( 2^n + 1 - n - 2 ) where n > 1, and the runsn scored in the k^th match are given byn k.2^n + 1 - k, where 1 <= k <= n. Find n

nn n S_n=sum _k=1^n k 2^n+1-k n =1 * 2^n+2 * 2^n-1+3 * 2^n-2+ * s * +(n-1) 2^2+n(2) n (1 / 2) S_n=1 * 2^n-1+2 * 2^n-2+3 * 2^n-3+ * s * +(n-1) 2+nnn nnn nSubtracting,nnnn n (1 / 2) S_n=1 * 2^n+1 * 2^n-1+1 * 2^n-2+ * s * +2-n=(2(2^n-1) / 2-1)-n n S_n=2(2^n+1-2-n)nn nnn nBut S_n=((n+1 / 4))(2^n+1-2-n) (given)nnnn n Therefore (n+1 / 4)(2^n+1-2-n)=2(2^n+1-2-n) n => (n+1 / 4)=2 => n=7nn nnn

If total number of runs scored in n matches is ( \fra... | Sequence & Series (Mathematics)

\n\n\begin{aligned}\n& S_n=\sum_{k=1}^n k 2^{n+1-k} \\\n& \quad=1 \cdot 2^n+2 \cdot 2^{n-1}+3 \cdot 2^{n-2}+\cdots \cdot+(n-1) 2^2+n(2) \\\n& \quad \frac{1}{2} S_n=1 \cdot 2^{n-1}+2 \cdot 2^{n-2}+3 \cdot 2^{n-3}+\cdots \cdot+(n-1) 2+n\n\n\end{aligned}\n\n

\n \nSubtracting,\n\n

\n\n\begin{aligned}\n& \frac{1}{2} S_n=1 \cdot 2^n+1 \cdot 2^{n-1}+1 \cdot 2^{n-2}+\cdots \cdot+2-n=\frac{2\left(2^n-1\right)}{2-1}-n \\\n& S_n=2\left(2^{n+1}-2-n\right)\n\n\end{aligned}\n\n

\n \nBut

S_n=\left(\frac{n+1}{4}\right)\left(2^{n+1}-2-n\right)

(given)\n\n

\n\n\begin{aligned}\n& \therefore \frac{n+1}{4}\left(2^{n+1}-2-n\right)=2\left(2^{n+1}-2-n\right) \\\n& \Rightarrow \frac{n+1}{4}=2 \Rightarrow n=7\n\n\end{aligned}\n\n

Explanation

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