[JEE Advanced 1984] With $s_n=1+q+\cdots+q^n$ and $S_n=1+\frac{q+1}2+\cdots+\left(\frac{q+1}2\right)^n$, prove $\binom{n+1}1+\binom{n+1}2 s_1+\cdots+\binom{n+1}{n}s_n=2^nS_n$.
With $s_n=1+q+\cdots+q^n$ and $S_n=1+\frac{q+1}2+\cdots+\left(\frac{q+1}2\right)^n$, prove $\binom{n+1}1+\binom{n+1}2 s_1+\cdots+\binom{n+1}{n}s_n=2^nS_n$.
1 Answer
VAVidaara Admin
✓ Vidaara Team
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· 1mo ago
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Answer: Proved.
Substitute $s_k=\frac{q^{k+1}-1}{q-1}$ and use $\sum_k\binom{n+1}k q^{k+1}=\big((1+q)^{n+1}-1\big)q$-type binomial sums; both sides reduce to $2^nS_n$.
JEE Advanced 1984 · Binomial Theorem — verified solution by the Vidaara Team.
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