The terms containing x^r y^s (for certain r a | vedclass

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(C) The general term in the expansion of $(x+y^2)^{13}$ is $T_{k+1} = \binom{13}{k} x^{13-k} (y^2)^k = \binom{13}{k} x^{13-k} y^{2k}$,where $0 \le k \

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$18$

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(C) The general term in the expansion of $(x+y^2)^{13}$ is $T_{k+1} = \binom{13}{k} x^{13-k} (y^2)^k = \binom{13}{k} x^{13-k} y^{2k}$,where $0 \le k \le 13$.Here,$r = 13-k$ and $s = 2k$.The general term in the expansion of $(x^2+y)^{14}$ is $T_{j+1} = \binom{14}{j} (x^2)^{14-j} y^j = \binom{14}{j} x^{28-2j} y^j$,where $0 \le j \le 14$.Here,$r = 28-2j$ and $s = j$.For the terms to be identical,we must have $13-k = 28-2j$ and $2k = j$.Substituting $j = 2k$ into the first equation: $13-k = 28-2(2k) \implies 13-k = 28-4k \implies 3k = 15 \implies k = 5$.Then $j = 2(5) = 10$.Since $k=5$ and $j=10$ are within the allowed ranges ($0 \le 5 \le 13$ and $0 \le 10 \le 14$),there is only $\alpha = 1$ such term.For this term,$r = 13-5 = 8$ and $s = 2(5) = 10$.The sum $r+s = 8+10 = 18$.Thus,$\alpha \sum (r+s) = 1 	imes 18 = 18$.

The terms containing $x^r y^s$ (for certain $r$ and $s$) are present in both the expansions of $(x+y^2)^{13}$ and $(x^2+y)^{14}$. If $\alpha$ is the number of such terms,then the sum $\alpha \sum_{r, s}(r+s) =$

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