Find a if the coefficients of x2 and x3 in the expansion of (3 + a x)9 are equal.

Asked by Pragya Singh | 1 year ago |  73

##### Solution :-

It is known that (r+1)th term $$(T_{r+1})$$ in the binomial expansion of (a+b)n is given by

$$(T_{r+1})=\;^nC_ra^{n-r}b^r$$ Assuming that x2 occurs in the (r+1)th term in the expansion of $$(3+ax)^9$$ we obtain

$$(T_{r+1})=\;^9C_r(3)^{9-r}(ax)^r$$

$$^9C_r(3)^{9-r}a^rx^r$$

Comparing the indices of x in x2 and in $$(T_{r+1})$$, we obtain r = 2

Thus, the coefficient of x2 is

$$^9C_r(3)^{9-2}a^2$$

$$= \dfrac{9!}{2!7!}(3)^7a^2=36(3)^7a^2$$

Assuming that x3 occurs in the (k+1)th term in the expansion of $$(3+ax)^9$$,we obtain

$$(T_{k+1})=\;^9C_r(3)^{9-k}(ax)^k$$

$$^9C_r(3)^{9-k}a^kx^k$$

Comparing the indices of x in x3 and in $$(T_{r+1})$$, we obtain k =3

Thus, the coefficient of x3 is

$$^9C_r(3)^{9-3}a^3$$

$$\dfrac{9!}{3!6!}(3)^6a^3=84(3)^6a^3$$

It is given that the coefficient of x2 and x3 are the same.

$$84(3)^6=36\times 3$$

$$a=\dfrac{36\times 3}{84}=\dfrac{104}{84}$$

$$a=\dfrac{9}{7}$$

Thus, the required value of a is $$\dfrac{9}{7}$$

Answered by Abhisek | 1 year ago

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