For m,n0, let (m,n)= _0^2t^m(1+3t)^ndt.

Question

For $m,n>0$, let $\alpha (m,n)=\int _{0}^{2}{t}^{m}(1+3t{)}^{n}dt$. If $11\alpha (10,6)+18\alpha (11,5)=p(14{)}^{6}$, then $p$ is equal to \_\_\_\_\_\_\_\_\_\_\_.

Answer: 32

Accepted Answer

We have, $\alpha (m,n)=\int _{0}^{2}{t}^{m}(1+3t{)}^{n}dt$

Also, $11\alpha (10,6)+18\alpha (11,5)=P(14{)}^{6}$

$\Rightarrow 11\int _{0}^{2}{t}^{10}(1+3t{)}^{6}dt+18\int _{0}^{2}{t}^{11}(1+3t{)}^{5}dt=P(14{)}^{6}$

Using integration by part for expression ${t}^{10}(1+3t{)}^{6}$

$\begin{gathered} \Rightarrow 11{[(1+3t{)}^{6}	imes \frac{{t}^{11}}{11}]}_{2}^{0}-\int _{0}^{2}6(1+3t{)}^{5}	imes 3	imes \frac{{t}^{11}}{11}dt] \ +18\int _{0}^{2}{t}^{11}(1+3t{)}^{5}dt=P(14{)}^{6} \end{gathered}$

$\begin{gathered} \Rightarrow {7}^{6}	imes {2}^{11}-18\int _{0}^{2}{t}^{11}(1+3t{)}^{5}dt+18\int _{0}^{2}{t}^{11}(1+3t{)}^{5}dt=P(14{)}^{6} \ \Rightarrow {7}^{6}	imes {2}^{11}=P(14{)}^{6}\Rightarrow (7	imes 2{)}^{6}	imes {2}^{5}=P(14{)}^{6} \ \Rightarrow P={2}^{5}=32 \end{gathered}$

Solutions · 1