Find the term independent of \(x\), \(x \neq 0\), in the expansion of \(\left(\dfrac{3x^{2}}{2}-\dfrac{1}{3x}\right)^{15}\).
\(\dfrac{1001}{2592}\)
If the term free from \(x\) in the expansion of \(\left(\sqrt{x}-\dfrac{k}{x^{2}}\right)^{10}\) is \(405\), find the value of \(k\).
\(k=\pm 3\)
Find the coefficient of \(x\) in the expansion of \((1-3x+7x^{2})(1-x)^{16}\).
\(-19\)
Find the term independent of \(x\) in the expansion of \(\left(3x-\dfrac{2}{x^{2}}\right)^{15}\).
\(-3003\cdot 3^{10}\cdot 2^{5}\) (i.e. \(-3003\,(3^{10})(2^{5})\))
Find the middle term(s) in the following expansions:
(i) \(\left(\dfrac{x}{a}-\dfrac{a}{x}\right)^{10}\)
(ii) \(\left(3x-\dfrac{x^{3}}{6}\right)^{9}\)
(i) middle term is the term with index \(k=5\): \(-252\).
(ii) the two middle terms are:
for \(k=4\): \(\dfrac{189}{8}x^{17}\),
for \(k=5\): \(-\dfrac{21}{16}x^{19}\).
Find the coefficient of \(x^{15}\) in the expansion of \((x-x^{2})^{10}\).
\(-252\)
Find the coefficient of \(\dfrac{1}{x^{17}}\) in the expansion of \(\left(x^{4}-\dfrac{1}{x^{3}}\right)^{15}\).
\(-1365\)
Find the sixth term of the expansion \(\left(\dfrac{1}{y^{2}}+\dfrac{1}{x^{3}}\right)^{n}\), if the binomial coefficient of the third term from the end is \(45\).
From \(\binom{n}{2}=45\) we get \(n=10\). Sixth term (\(k=5\)) is \(\displaystyle \binom{10}{5}\,\dfrac{1}{y^{10}}\dfrac{1}{x^{15}}=252\,x^{-15}y^{-10}\).
Find the value of \(r\), if the coefficients of the \((2r+4)\)-th and \((r-2)\)-th terms in the expansion of \((1+x)^{18}\) are equal.
\(r=6\)
If the coefficients of the second, third and fourth terms in the expansion of \((1+x)^{2n}\) are in A.P., show that \(2n^{2}-9n+7=0\).
Let the coefficients be \(C_{1}=\binom{2n}{1}=2n\), \(C_{2}=\binom{2n}{2}=n(2n-1)\), \(C_{3}=\binom{2n}{3}=\dfrac{(2n)(2n-1)(2n-2)}{6}\). A.P. condition: \(2C_{2}=C_{1}+C_{3}\). Substituting and simplifying gives \(2n^{2}-9n+7=0\).
Find the coefficient of \(x^{4}\) in the expansion of \((1+x+x^{2}+x^{3})^{11}\).
Number of weak compositions of 4 into 11 parts (each \(\le 3\)) equals \(\binom{4+11-1}{11-1}=\binom{14}{10}=\binom{14}{4}=1001\).