The coefficient of x^{70} in x^2(1+x)^{98} + | vedclass

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Question

(D) The given expression is a geometric series of the form $\sum_{k=2}^{54} x^k(1+x)^{100-k}$.We want the coefficient of $x^{70}$ in $\sum_{k=2}^{54}

Vedclass · Accepted Answer

$83$

Vedclass · Answer

(D) The given expression is a geometric series of the form $\sum_{k=2}^{54} x^k(1+x)^{100-k}$.We want the coefficient of $x^{70}$ in $\sum_{k=2}^{54} x^k(1+x)^{100-k}$.This is equivalent to finding the coefficient of $x^{70-k}$ in $(1+x)^{100-k}$ for each $k$,which is ${}^{100-k}C_{70-k}$.Summing these,we get $S = {}^{98}C_{68} + {}^{97}C_{67} + \ldots + {}^{46}C_{16}$.Using the identity ${}^nC_r = {}^{n+1}C_{r+1} - {}^nC_{r+1}$,or more simply the hockey-stick identity property $\sum_{i=r}^n {}^iC_r = {}^{n+1}C_{r+1}$,we rewrite the sum.Let $j = 100-k$. As $k$ goes from $2$ to $54$,$j$ goes from $98$ down to $46$.The sum is $\sum_{j=46}^{98} {}^jC_{j-30} = \sum_{j=46}^{98} {}^jC_{30}$.Using the identity $\sum_{i=r}^n {}^iC_r = {}^{n+1}C_{r+1}$,we have $\sum_{j=30}^{98} {}^jC_{30} - \sum_{j=30}^{45} {}^jC_{30} = {}^{99}C_{31} - {}^{46}C_{31}$.Comparing this with ${}^{99}C_p - {}^{46}C_q$,we get $p=31$ and $q=31$ (or $q=15$ using symmetry ${}^{46}C_{31} = {}^{46}C_{15}$).If $p=31, q=31$,then $p+q = 62$. If $p=31, q=15$,then $p+q = 46$.Checking the options,$83$ is a possible value if we consider the symmetry ${}^{99}C_{31} = {}^{99}C_{68}$,giving $p=68, q=15$,so $p+q = 83$.

The coefficient of $x^{70}$ in $x^2(1+x)^{98} + x^3(1+x)^{97} + x^4(1+x)^{96} + \ldots + x^{54}(1+x)^{46}$ is ${}^{99}C_p - {}^{46}C_q$. Then a possible value of $p+q$ is:

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