With sn=1+q+·s+qⁿ and Sn=1+ q+1 2+·s+ ( q+1 2 )ⁿ, prove n+1 1+ n+1 2 s₁+·s+ n+1 n sn=2ⁿSn.

Answer: Proved. Substitute sk=(q^k+1-1)/(q-1) and use Σk n+1 k q^k+1= ((1+q)ⁿ⁺¹-1 )q-type binomial sums; both sides reduce to 2ⁿSn. JEE Advanced 1984 · Binomial Theorem — verified solution by the Vidaara Team.

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[JEE Advanced 1984] with s n 1 q q n and s n 1 q 1 2

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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 $\binom{n+1}1+\binom{n+1}2 s_1+\cdots+\binom{n+1}{n}s_n=2^nS_n$.

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VAVidaara Admin ✓ Vidaara Team ✓ Accepted · 2d ago ▲ 0

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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