If C_r = ^{100}C_r,then the value of 1 cdot C | vedclass

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(B) Let $S = 1 \cdot C_0^2 - 2 \cdot C_1^2 + 3 \cdot C_2^2 - 4 \cdot C_3^2 + \dots + 101 \cdot C_{100}^2$. Using the property $C_r = C_{n-r}$,we can w

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$51 \cdot ^{100}C_{50}$

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(B) Let $S = 1 \cdot C_0^2 - 2 \cdot C_1^2 + 3 \cdot C_2^2 - 4 \cdot C_3^2 + \dots + 101 \cdot C_{100}^2$. Using the property $C_r = C_{n-r}$,we can write the series in reverse order: $S = 101 \cdot C_{100}^2 - 100 \cdot C_{99}^2 + 99 \cdot C_{98}^2 - \dots + 1 \cdot C_0^2$. Adding the two expressions for $S$: $2S = (1+101)C_0^2 - (2+100)C_1^2 + (3+99)C_2^2 - \dots + (101+1)C_{100}^2$. $2S = 102(C_0^2 - C_1^2 + C_2^2 - C_3^2 + \dots + C_{100}^2)$. The sum $C_0^2 - C_1^2 + C_2^2 - \dots + C_{100}^2$ is the coefficient of $x^{100}$ in the expansion of $(1-x^2)^{100}$,which is $(-1)^{50} \cdot ^{100}C_{50} = ^{100}C_{50}$. Thus,$2S = 102 \cdot ^{100}C_{50}$,which simplifies to $S = 51 \cdot ^{100}C_{50}$.

If $C_r = ^{100}C_r$,then the value of $1 \cdot C_0^2 - 2 \cdot C_1^2 + 3 \cdot C_2^2 - 4 \cdot C_3^2 + \dots + 101 \cdot C_{100}^2$ is equal to:

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