[JEE Advanced 2005] Runs in the $k$-th of $n$ matches are $k\cdot2^{\,n+1-k}$ and the total is $\frac{n+1}4(2^{\,n+1}-n-2)$. Find $n$.
Runs in the $k$-th of $n$ matches are $k\cdot2^{\,n+1-k}$ and the total is $\frac{n+1}4(2^{\,n+1}-n-2)$. Find $n$.
1 Answer
VAVidaara Admin
✓ Vidaara Team
✓ Accepted
· 1mo ago
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Answer: $n=7$.
$\sum_{k=1}^n k\,2^{\,n+1-k}=2^{\,n+2}-2n-4$; equating to the given total gives $n=7$ (both sides equal $494$).
JEE Advanced 2005 · Permutations and Combinations — verified solution by the Vidaara Team.
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