If the coefficients of x

If the coefficients of x7 and x8 in (2+x3)n2+x3n are equal, then n is

(a) 56

(b) 55

(c) 45

(d) 15


In $(a+x)^{n}$

$T_{r+1}={ }^{n} C_{r} a^{n-r} x^{r}$

$\therefore \operatorname{In}\left(2+\frac{x}{3}\right)^{n}$

Coefficient of x7 is

$T_{8}={ }^{n} C_{7}(2)^{n-7}\left(\frac{x}{3}\right)^{7}$

$T_{8}={ }^{n} C_{7} \frac{2^{\text {n- }-} x^{7}}{3^{7}}$

and coefficient of x8 is

$T_{9}={ }^{n} C_{8}(2)^{n-8}\left(\frac{x}{3}\right)^{8}$

$T_{9}={ }^{n} C_{8} \frac{2^{n-8} x^{8}}{3^{8}}$

Since $T_{8}=T_{9}$ i.e coefficient of $T_{8}=$ coefficient of $T_{9}$

$\Rightarrow^{n} C_{7} \frac{2^{n-7}}{3^{7}}=\frac{{ }^{n} C_{8} 2^{n-8}}{3^{8}}$

$\Rightarrow \frac{n ! 2^{n-7}}{7 !(n-7) ! \times 3^{7}}=\frac{n ! \times 2^{n-8}}{8 !(n-8) ! \times 3^{8}}$

i. e. $\frac{8 !}{7 !} \frac{(n-8) !}{(n-7)(n-8) !}=\frac{2^{n-8}}{3 \times 2^{n-7}}$

i.e. $\frac{8}{n-7}=\frac{1}{6} \Rightarrow 48=n-7$

i. e. $n=55$

Hence, the correct answer is option B.






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