# Maths Past Paper Question

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Hey guys If anybody could solve and explain this question to me, I'd really appreciate it.

Given that (1+2/3x)n (3+nx)2 = 9 + 84x+..., find the value of n.

Anyone?

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Hey guys If anybody could solve and explain this question to me, I'd really appreciate it.

Given that (1+2/3x)n (3+nx)2 = 9 + 84x+..., find the value of n.

I got n = 7, is it the correct answer? Here's the working:

By binomial theorem:

[E1] $\left( 1 + \frac{2}{3} x\right)^n = \sum_{k=0}^n {n \choose k} 1^{n-k} \left(\frac{2}{3} x\right)^k = \sum_{k=0}^n {n \choose k} \left(\frac{2}{3} x\right)^k$

Now, expanding (3+nx)^2, we have:

[E2] $\left(3+nx\right)^2 = 9 + 6nx +n^2x^2$

Notice that equation [E2] doesn't have any 'k', so we can just combine that in the sum in equation [E1]:

[E3] $\left( 1 + \frac{2}{3} x\right)^n \left(3+nx\right)^2 = \sum_{k=0}^n \left{ {n \choose k} \left(\frac{2}{3} x\right)^k \left( 9+6n+n^2x^2\right) \right}$

Expanding equation [E3] to get:

[E4] $\left( 1 + \frac{2}{3} x\right)^n \left(3+nx\right)^2 = \sum_{k=0}^n \left\{ 9{n \choose k} \left(\frac{2}{3}\right)^k x^k + 6n{n \choose k} \left(\frac{2}{3}\right)^k x^{k+1} + n^2{n \choose k} \left(\frac{2}{3}\right)^k x^{k+2} \right\}$

From equation [E4], there are only 2 ways that you can get x^1.

The first way is to let $k = 0$ for the term $6n{n \choose k} \left(\frac{2}{3}\right)^k x^{k+1}$

• Thus, $6n{n \choose k} \left(\frac{2}{3}\right)^k x^{k+1} = 6n{n \choose 0} \left(\frac{2}{3}\right)^0 x^{0+1} = 6n x^{1}$

The second way is to let $k = 1$ for the term $9{n \choose k} \left(\frac{2}{3}\right)^k x^k$

• Thus, $9{n \choose k} \left(\frac{2}{3}\right)^k x^k = 9{n \choose 1} \left(\frac{2}{3}\right)^1 x^1 = 6n x^1$

Now, adding the 2 ways together, we get:

$6nx^1 + 6nx^1 = 12nx = 84x \Leftrightarrow 12n = 84 \Leftrightarrow n = 7$

Hope this helps

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Hey guys If anybody could solve and explain this question to me, I'd really appreciate it.

Given that (1+2/3x)n (3+nx)2 = 9 + 84x+..., find the value of n.

I got n = 7, is it the correct answer? Here's the working:

By binomial theorem:

[E1] $\left( 1 + \frac{2}{3} x\right)^n = \sum_{k=0}^n {n \choose k} 1^{n-k} \left(\frac{2}{3} x\right)^k = \sum_{k=0}^n {n \choose k} \left(\frac{2}{3} x\right)^k$

Now, expanding (3+nx)^2, we have:

[E2] $\left(3+nx\right)^2 = 9 + 6nx +n^2x^2$

Notice that equation [E2] doesn't have any 'k', so we can just combine that in the sum in equation [E1]:

[E3] $\left( 1 + \frac{2}{3} x\right)^n \left(3+nx\right)^2 = \sum_{k=0}^n \left{ {n \choose k} \left(\frac{2}{3} x\right)^k \left( 9+6n+n^2x^2\right) \right}$

Expanding equation [E3] to get:

[E4] $\left( 1 + \frac{2}{3} x\right)^n \left(3+nx\right)^2 = \sum_{k=0}^n \left\{ 9{n \choose k} \left(\frac{2}{3}\right)^k x^k + 6n{n \choose k} \left(\frac{2}{3}\right)^k x^{k+1} + n^2{n \choose k} \left(\frac{2}{3}\right)^k x^{k+2} \right\}$

From equation [E4], there are only 2 ways that you can get x^1.

The first way is to let $k = 0$ for the term $6n{n \choose k} \left(\frac{2}{3}\right)^k x^{k+1}$

• Thus, $6n{n \choose k} \left(\frac{2}{3}\right)^k x^{k+1} = 6n{n \choose 0} \left(\frac{2}{3}\right)^0 x^{0+1} = 6n x^{1}$

The second way is to let $k = 1$ for the term $9{n \choose k} \left(\frac{2}{3}\right)^k x^k$

• Thus, $9{n \choose k} \left(\frac{2}{3}\right)^k x^k = 9{n \choose 1} \left(\frac{2}{3}\right)^1 x^1 = 6n x^1$

Now, adding the 2 ways together, we get:

$6nx^1 + 6nx^1 = 12nx = 84x \Leftrightarrow 12n = 84 \Leftrightarrow n = 7$

Hope this helps

Thank you so much!! I need your brain for the exam tomorrow

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