If the non-zero coefficient of the (2r + 4)^{ | vedclass

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(A) The general term of $(1 + x)^{18}$ is $T_{k+1} = ^{18}C_k x^k$. The $(2r + 4)^{th}$ term is $T_{2r+4}$,where $k = 2r + 3$. Its coefficient is $^{1

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$3$

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(A) The general term of $(1 + x)^{18}$ is $T_{k+1} = ^{18}C_k x^k$. The $(2r + 4)^{th}$ term is $T_{2r+4}$,where $k = 2r + 3$. Its coefficient is $^{18}C_{2r+3}$. The $(r - 2)^{th}$ term is $T_{r-2}$,where $k = r - 3$. Its coefficient is $^{18}C_{r-3}$. Given the condition: $^{18}C_{2r+3} > ^{18}C_{r-3}$. For the binomial coefficients to be defined,we must have $0 \le 2r+3 \le 18$ and $0 \le r-3 \le 18$. From $2r+3 \ge 0$,$r \ge -1.5$. From $2r+3 \le 18$,$r \le 7.5$. From $r-3 \ge 0$,$r \ge 3$. From $r-3 \le 18$,$r \le 21$. Combining these,$3 \le r \le 7.5$. Thus $r \in \{3, 4, 5, 6, 7\}$. Testing the inequality $^{18}C_{2r+3} > ^{18}C_{r-3}$: If $r=3$: $^{18}C_9 > ^{18}C_0$ ($48620 > 1$,True). If $r=4$: $^{18}C_{11} > ^{18}C_1$ ($31824 > 18$,True). If $r=5$: $^{18}C_{13} > ^{18}C_2$ ($8568 > 153$,True). If $r=6$: $^{18}C_{15} > ^{18}C_3$ ($816 > 816$,False). If $r=7$: $^{18}C_{17} > ^{18}C_4$ ($18 > 3060$,False). The possible integral values for $r$ are $3, 4, 5$. The total number of values is $3$.

If the non-zero coefficient of the $(2r + 4)^{th}$ term is greater than the non-zero coefficient of the $(r - 2)^{th}$ term in the expansion of $(1 + x)^{18}$,then the number of possible integral values of $r$ is:

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