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i) Find in terms of the non zero constant $k$, the first four terms on the expansion $(k+x)^n$ in ascending powers of $x$

ii) Given that the cooefficients of the $x^2$ and $x^{3}$ are equal, find the value of $k$.

I really neeed help on these so any help is highly appreciated!

No clue how to start. :(

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    $\begingroup$ Since you are new, I want to give some advice about the site: To get the best possible answers, you should explain what your thoughts on the problem are so far. That way, people won't tell you things you already know, and they can write answers at an appropriate level; also, people tend to be more willing to help you if you show that you've tried the problem yourself. $\endgroup$ – Zev Chonoles Jun 19 '13 at 19:20
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    $\begingroup$ Do you really mean $x^{-3}$, or do you mean $x^3$? Because there is no term $x^{-3}$ in the expansion. $\endgroup$ – Thomas Andrews Jun 19 '13 at 19:23
  • $\begingroup$ Oh sorry, your right it's x^3 $\endgroup$ – Sachin Jun 19 '13 at 19:23
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HINT: At this point you’re expected to know the binomial expansion:

$$(k+x)^n=\sum_{i=0}^n\binom{n}ik^{n-i}x^i\;.$$

The first four terms are the terms for $i=0,1,2,3$:

$$\binom{n}0k^n,\quad\binom{n}1k^{n-1}x,\quad\binom{n}2k^{n-2}x^2,\quad\text{and}\quad\binom{n}3k^{n-3}x^3\;.$$

Since the coefficients of $x^2$ and $x^3$ are equal, you know that

$$\binom{n}2k^{n-2}=\binom{n}3k^{n-3}\;.$$

Divide out the common factors and expand the binomial coefficients into fairly simple fractions, and you’ll be able to solve for $k$ as a fairly simple function of $n$.

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Hint: use binomial coefficients.

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