[JEE Advanced 1979] Given $\sum_{r=1}^{2n}rC_rx^{r-1}=2n(1+x)^{2n-1}$ with $C_r=\binom{2n}r$, prove $C_1^2-2C_2^2+3C_3^2-\cdots-2nC_{2n}^2=(-1)^n n\binom{2n}n$.
Given $\sum_{r=1}^{2n}rC_rx^{r-1}=2n(1+x)^{2n-1}$ with $C_r=\binom{2n}r$, prove $C_1^2-2C_2^2+3C_3^2-\cdots-2nC_{2n}^2=(-1)^n n\binom{2n}n$.
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✓ Vidaara Team
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Answer: Proved $=(-1)^n n\binom{2n}n$.
Multiply the given derivative identity by the expansion of $(1-x)^{2n}$ (or its derivative) and compare the coefficient of $x^{2n-1}$; the alternating weighted sum of squares emerges as $(-1)^n n\binom{2n}n$.
JEE Advanced 1979 · Binomial Theorem — verified solution by the Vidaara Team.
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