1
$\begingroup$

The coefficient of $x^2$ in the expansion of $\left(1 + \frac x5\right)^n$, where $n$ is a positive integer, is $\frac 35$ .

$(i)$ Find the value of $n$.

$(ii)$ Using this value of $n$, find the term independent of $x$ in the expansion of $\left(1 + \frac x5\right)^n \times \left(2- \frac 3x\right)^2$

For part $(i)$ I used the Binomial theorem and got the result where $n= 6$. Had no bigger issues with solving for $n$.

I do, however, struggle with part $(ii)$. I am not quite sure what I am supposed to use/do here. Do I also use the binomial theorem? I tried that and got nowhere since I do not know what "$r$" or "$n$" value to use since there are two different powers of binomial.

Any help would be appreciated.

$\endgroup$
1
$\begingroup$

For a general $n$, the constant term in $$f(x) = (1 + x/5)^n (2-3/x)^2$$ can be found by observing first that $$(2-3/x)^2 = 2 - 12x^{-1} + 9x^{-2}.$$ Then we see that the constant term of $f$ is given by $$1 \cdot 2 + \binom{n}{1} (x/5)(-12x^{-1}) + \binom{n}{2}(x/5)^2(9x^{-2}),$$ since these are the only terms for which the power of $x$ will be zero.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.