[JEE Advanced 1992] if sum r 0 2n a r x 2 r sum r 0 2n
If $\sum_{r=0}^{2n}a_r(x-2)^r=\sum_{r=0}^{2n}b_r(x-3)^r$ and $a_k=1$ for all $k\ge n$, show that $b_n=\binom{2n+1}{n+1}$.
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VAVidaara Admin
✓ Vidaara Team
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· 2d ago
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Answer: Proved.
Since $b_n$ is the coefficient of $(x-3)^n$, expand $(x-2)^r=((x-3)+1)^r$; collecting the $(x-3)^n$ terms with $a_k=1$ for $k\ge n$ gives $b_n=\sum_{r\ge n}\binom rn=\binom{2n+1}{n+1}$.
JEE Advanced 1992 · Binomial Theorem — verified solution by the Vidaara Team.
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