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7.Binomial Theorem
normal
If sum of the coefficient of the first, second and third terms of the expansion of ${\left( {{x^2} + \frac{1}{x}} \right)^m}$ is $46$, then the coefficient of the term that doesnot contain $x$ is :-
A
$84$
B
$92$
C
$98$
D
$106$
Solution
It is given $^m{C_0} + {\,^m}{C_1} + {\,^m}{C_2} = 46$
$\Rightarrow 2 m+m(m-1)=90$
$\Rightarrow m^{2}+m-90=0$
$\Rightarrow m=9$ as $m>0$
Now $(\mathrm{r}+1)^{\text {th }}$ term of $\left(\mathrm{x}^{2}+\frac{1}{\mathrm{x}}\right)^{\mathrm{m}}$ is
$ = {\,^m}{C_r}{\left( {{x^2}} \right)^{m – r}}{\left( {\frac{1}{x}} \right)^r}$
For this to be independent of $x$
$2 m-3 r=0 \Rightarrow r=6$
Standard 11
Mathematics
