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The number of positive integers \mathrm{k} such that the constant term in the binomial expansion of \mathrm{\left(2 x^{3}+\frac{3}{x^{k}}\right)^{12}, x \neq 0 \: is\: 2^{8} \cdot \ell,} where \mathrm{\ell } is an odd integer, is __________.

Option: 1

2


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

\mathrm{T_{r+1}={ }^{12} C_{r}\left(2 x^{3}\right)^{12-r}\left(\frac{3}{x^{k}}\right)^{r}}

\mathrm{={ }^{12} C_{r} \cdot 2^{12-r} \cdot 3^{r} \cdot x^{36-3 r-k r}}

\mathrm{\text{For constant term}, 36-r(3+k)=0}
\mathrm{\Rightarrow r=\frac{36}{3+k} ; k \in N \text {. }}

\mathrm{So\; \; 3+k=\{4,6,9,12,18,36\}.}
\mathrm{\Rightarrow k=\{1,3,6,9,15,33\} \text {. }}

\mathrm{Corresponding\; \; r=\{9,6,4,3,2,1\}.}

\mathrm{(i) \; \; T_{10}={ }^{12} C_{9} \cdot 2^{3} \cdot 3^{9}}
\mathrm{No.\: of \; 25\; in \; \; \frac{12 !}{9 ! 3 !} are \left(\left[\frac{12}{2}\right]+\left[\frac{12}{4}\right]+\left[\frac{12}{2}\right]+\cdots\right)}
\mathrm{-\left(\left[\frac{9}{2}\right]+\left[\frac{9}{4}\right]+\left[\frac{9}{2}\right]+\cdots\right)-\left(\left[\frac{3}{2}\right]+\cdots\right)}

 \mathrm{=10-7-1=2}

\mathrm{So \; No.\; \: of \; \; 2 s=3+2=5 x}


\mathrm{(ii) \; \; T_{7}={ }^{12} C_{6} \cdot 2^{6} \cdot 3^{6}}
\mathrm{No. \: of \; 2s\; in \; \frac{12 !}{6 ! 6 !}=10-4-4=2.}
\mathrm{So \; No.\; of \; 2 s=6+2=8}


\mathrm{(iii)\; \; T_{5}={ }^{12} C_{4} \times 2^{8} \times 3^{4}}
\mathrm{No.\: of \; 2 s\; \; in\: \frac{12!}{8!4!}=10-(4+2+1)-(2+1)=0}
\mathrm{No.\: of \; 2 s\; \; = 8+0= 8}

\mathrm{(iv)\: \; r= 3,2,1} are rejected as No.of  \mathrm{2s> 8.}

So No.of positive integers \mathrm{k= 2.}
Which are 3 & 6
 

Posted by

Ramraj Saini

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