If sumlimits_{i = 0}^4 {^{4 + i}} {C_i} + sum | vedclass

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(C) The given expression is $S = \sum_{i=0}^4 {^{4+i}C_i} + \sum_{j=6}^9 {^{3+j}C_j}$.Expanding the first sum: ${^4C_0} + {^5C_1} + {^6C_2} + {^7C_3}

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$(x - y)$ and $(x + y)$ will always be co-prime numbers.

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(C) The given expression is $S = \sum_{i=0}^4 {^{4+i}C_i} + \sum_{j=6}^9 {^{3+j}C_j}$.Expanding the first sum: ${^4C_0} + {^5C_1} + {^6C_2} + {^7C_3} + {^8C_4}$.Using the identity ${^nC_r} + {^nC_{r-1}} = {^{n+1}C_r}$,we note ${^4C_0} = {^5C_0} = 1$.So,${^5C_0} + {^5C_1} = {^6C_1}$,then ${^6C_1} + {^6C_2} = {^7C_2}$,then ${^7C_2} + {^7C_3} = {^8C_3}$,then ${^8C_3} + {^8C_4} = {^9C_4}$.Now,the expression becomes ${^9C_4} + {^9C_6} + {^{10}C_7} + {^{11}C_8} + {^{12}C_9}$.Using ${^9C_4} = {^9C_5}$,we get ${^9C_5} + {^9C_6} = {^{10}C_6}$.Then ${^{10}C_6} + {^{10}C_7} = {^{11}C_7}$.Then ${^{11}C_7} + {^{11}C_8} = {^{12}C_8}$.Then ${^{12}C_8} + {^{12}C_9} = {^{13}C_9}$.Thus,${^xC_y} = {^{13}C_9} = {^{13}C_4}$.Here $x = 13$ (which is prime),so $y = 9$ or $y = 4$.If $y = 9$,$x-y = 4$ and $x+y = 22$. If $y = 4$,$x-y = 9$ and $x+y = 17$.Option $C$ is incorrect because $4$ and $22$ are not co-prime.

List-$I$	List-$II$
$(A)$ $a_0 + a_2 + \ldots + a_{2n}$	$(I)$ $n \cdot 3^{n-1}$
$(B)$ $a_1 + a_3 + \ldots + a_{2n-1}$	$(II)$ $n \cdot 3^n$
$(C)$ $a_1 + 2a_2 + 3a_3 + \ldots + 2n a_{2n}$	$(III)$ $\frac{1}{2}(3^n + 1)$
	$(IV)$ $\frac{1}{2}(3^n - 1)$

If $\sum\limits_{i = 0}^4 {^{4 + i}} {C_i} + \sum\limits_{j = 6}^9 {^{3 + j}} {C_j} = {\,^x}{C_y}$ ($x$ is a prime number),then which one of the following is incorrect?

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