Let (1 + x)^m = C_0 + C_1x + C_2x^2 + C_3x^3 | vedclass

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(A) We know that the sum of products of binomial coefficients is given by $\sum_{r=0}^{m-k} C_r C_{r+k} = {}^{2m}C_{m-k}$.For $k=2$,we have $\sum_{r=0

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$A \ge {}^{2m}C_{m-2}$

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(A) We know that the sum of products of binomial coefficients is given by $\sum_{r=0}^{m-k} C_r C_{r+k} = {}^{2m}C_{m-k}$.For $k=2$,we have $\sum_{r=0}^{m-2} C_r C_{r+2} = C_0C_2 + C_1C_3 + . . . + C_{m-2}C_m = {}^{2m}C_{m-2}$.Given $A = C_1C_3 + C_2C_4 + . . . + C_{m-2}C_m$,we can write:$A = (C_0C_2 + C_1C_3 + . . . + C_{m-2}C_m) - C_0C_2$.Since $C_0 = 1$ and $C_2 = {}^mC_2$,we get $A = {}^{2m}C_{m-2} - {}^mC_2$.Thus,option $C$ is true.Since ${}^mC_2 > 0$,it follows that $A Also,$\sum_{r=0}^m C_r^2 = {}^{2m}C_m$. Since $A Therefore,the false statement is $A \ge {}^{2m}C_{m-2}$.

Let $(1 + x)^m = C_0 + C_1x + C_2x^2 + C_3x^3 + . . . + C_mx^m$,where $C_r = {}^mC_r$ and $A = C_1C_3 + C_2C_4 + C_3C_5 + . . . + C_{m-2}C_m$. Which of the following is false?

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