If the coefficients of x and x2 in the expansion of (1 + x)p (1 – x)q, p, q≤15, are – 3 and – 5 respectively, then coefficient of x3 is equal to ______.
Coefficient of x in (1 + x)p (1 – x)q
\(^{−p}C_0\ ^{q}C_1+ ^{p}C_1\ ^qC_0=−3\)
\(⇒ p−q=−3\)
Coefficient of x2 in (1 + x)p (1 – x)q
\(^pC_0\ ^qC_2− ^pC_1\ ^qC_1 + ^pC_2\ ^qC_0=−5\)
\(\frac{q(q−1)}{2}−pq+\frac{p(p−1)}{2}=−5\)
\(\frac{q^2−q}{2}−(q−3)q+\frac{(q−3)(q−4)}{2}=−5\)
⇒ q = 11, p = 8
Coefficient of x3 in (1 + x)8 (1 – x)11 is
\(=^{−11}C_3+ ^8C_1 ^{11}C_2− ^8C_2 ^{11}C_1+ ^8C_3\)
=23
So, the answer is 23.
\[ \left( \frac{1}{{}^{15}C_0} + \frac{1}{{}^{15}C_1} \right) \left( \frac{1}{{}^{15}C_1} + \frac{1}{{}^{15}C_2} \right) \cdots \left( \frac{1}{{}^{15}C_{12}} + \frac{1}{{}^{15}C_{13}} \right) = \frac{\alpha^{13}}{{}^{14}C_0 \, {}^{14}C_1 \cdots {}^{14}C_{12}} \]
Then \[ 30\alpha = \underline{\hspace{1cm}} \]
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