Given Σr=1²ⁿrCrx^r-1=2n(1+x)²ⁿ⁻¹ with Cr= 2n r, prove C₁²-2C₂²+3C₃²-·s-2nC2n²=(-1)ⁿ n 2n n.

Answer: Proved =(-1)ⁿ n 2n n. Multiply the given derivative identity by the expansion of (1-x)²ⁿ (or its derivative) and compare the coefficient of x²ⁿ⁻¹; the alternating weighted sum of squares emerges as (-1)ⁿ n 2n n. JEE Advanced 1979 · Binomial Theorem — verified solution by the Vidaara Team.

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[JEE Advanced 1979] given sum r 1 2n rc rx r 1 2n 1 x 2n 1

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

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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