Question

Find the value of the constant K when:

Answer 1

The coefficient of x^2 in the expansion of \[(k + \frac{ 1 }{ 3 } x)^5 = 30 \]

Answer 2

One way is to expand the expression but I won't advise it.

Answer 3

What I would suggest is using this instead:

For the expansion of \((x+a)^n\), the rth term is given by:

\[\large{\boxed{T_{r} = ^nC_{r-1}x^ra^{n-r}}}\]

Now, first find out the term in which power of x is 2.

Answer 4

This is what I got, not too sure about how to solve for r. Sometimes I get it, sometimes I don't.

5-2r = 2
-2r = -3
r = 3/2

Answer 5

Okay lets see:

First of all I am going to simplify the expansion a little bit:

\[\large{(k + \cfrac{1}{3}x)^5 = \cfrac{1}{3^5}(3k + x)^5}\]


Now, since, the term you are asking for has power of x = 2

Thus, r = 2

Also, from the simplified expansion,

a = 3k

n = 5

So, the term would be:

\[\large{T_2 = \cfrac{1}{3^5}\times ^5C_1 x^2 (3k)^{5-2}}\]

Now, the coefficient part is clearly:

\[\large{\cfrac{1}{3^5}\times^5C_1\times(3k)^3}\] (I am multiplying by 1/3^5 as it was multiplied by the whole expansion)

But you are given that this is equal to 30. So,

\[\large{\cfrac{1}{3^5}\times ~^5C_1\times (3k)^3 = 30}\]

So, use this equation to find out k.

Answer 6

Not sure if I'm right:
\[\frac{ 1 }{3^5 } \times ^5C_{1} \times (3k)^3 = 30\]
\[\frac{ 5 \times 27k^3 }{3^5 } = 30\]
\[135k^3 = 7290\]
\[k^3 = 54\]
\[k = 3.78\]

Answer 7

You are correct (as far as I think)

Answer 8

Given answer is 3 though, it could be the wrong answer - will check with my lecturer tomorrow. Thanks for the help! :)