If sum_{i=0}^4 {^{4+i}}C_i + sum_{j=6}^9 {^{3 | 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: ${^4}C_0 + {^5}C_1 + {^6}C_2 + {^7}C_3

Vedclass · Accepted Answer

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

Vedclass · Answer

(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: ${^4}C_0 + {^5}C_1 + {^6}C_2 + {^7}C_3 + {^8}C_4$.Using the identity ${^n}C_r + {^n}C_{r-1} = {^{n+1}}C_r$,we note ${^4}C_0 = {^5}C_0 = 1$.So,${^5}C_0 + {^5}C_1 = {^6}C_1$,then ${^6}C_1 + {^6}C_2 = {^7}C_2$,then ${^7}C_2 + {^7}C_3 = {^8}C_3$,and finally ${^8}C_3 + {^8}C_4 = {^9}C_4$.Now,the total sum is ${^9}C_4 + {^9}C_6 + {^{10}}C_7 + {^{11}}C_8 + {^{12}}C_9$.Since ${^9}C_4 = {^9}C_5$,we have ${^9}C_5 + {^9}C_6 = {^{10}}C_6$.Then ${^{10}}C_6 + {^{10}}C_7 = {^{11}}C_7$,then ${^{11}}C_7 + {^{11}}C_8 = {^{12}}C_8$,and finally ${^{12}}C_8 + {^{12}}C_9 = {^{13}}C_9$.Thus,${^x}C_y = {^{13}}C_9$ or ${^{13}}C_4$. Here $x = 13$,which is prime.Possible pairs $(x, y)$ are $(13, 9)$ and $(13, 4)$.For $(13, 9)$: $x-y = 4, x+y = 22$. For $(13, 4)$: $x-y = 9, x+y = 17$.Option $C$ states they are always co-prime,but for $(13, 9)$,$gcd(4, 22) = 2 
eq 1$. Thus,$C$ is incorrect.

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

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