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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. 

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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] gif.latex? \left( 1 + \frac{2}{3} x\righ

 

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

[E2] gif.latex? \left(3+nx\right)^2 = 9 + 6nx

 

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

[E3] gif.latex? \left( 1 + \frac{2}{3} x\righ

 

Expanding equation [E3] to get:

 

[E4] gif.latex? \left( 1 + \frac{2}{3} x\righ

 

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

 

The first way is to let gif.latex? k = 0 for the term gif.latex? 6n{n \choose k} \left(\frac{2

  • Thus, gif.latex?6n{n \choose k} \left(\frac{2}

The second way is to let gif.latex? k = 1 for the term gif.latex? 9{n \choose k} \left(\frac{2}

  • Thus, gif.latex? 9{n \choose k} \left(\frac{2}

Now, adding the 2 ways together, we get:

gif.latex? 6nx^1 + 6nx^1 = 12nx = 84x \L

 

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] gif.latex? \left( 1 + \frac{2}{3} x\righ

 

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

[E2] gif.latex? \left(3+nx\right)^2 = 9 + 6nx

 

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

[E3] gif.latex? \left( 1 + \frac{2}{3} x\righ

 

Expanding equation [E3] to get:

 

[E4] gif.latex? \left( 1 + \frac{2}{3} x\righ

 

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

 

The first way is to let gif.latex? k = 0 for the term gif.latex? 6n{n \choose k} \left(\frac{2

  • Thus, gif.latex?6n{n \choose k} \left(\frac{2}

The second way is to let gif.latex? k = 1 for the term gif.latex? 9{n \choose k} \left(\frac{2}

  • Thus, gif.latex? 9{n \choose k} \left(\frac{2}

Now, adding the 2 ways together, we get:

gif.latex? 6nx^1 + 6nx^1 = 12nx = 84x \L

 

Hope this helps :)

 

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

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