# Thread: Binomial Theorem HSC 1985 Q6

1. ## Binomial Theorem HSC 1985 Q6

Could anyone help me with part ii of this question?

I finished part i and got 14. Im guessing part ii has something to do with using the 3+2x and maybe n=10 and some sort of sub for x. Part from that I'm lost.

Any help appreciated, thanks.

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2. ## Re: Binomial Theorem HSC 1985 Q6

$(x + 3)^{10} = \binom {10} 0 x^{10}+ \binom {10} 1 x^9\cdot3^1 +\binom {10} 2 x^8 \cdot 3^2 + \cdots + \binom {10} 9 x^1 \cdot 3^9 + \binom {10} {10} x^0 \cdot 3^{10}$

Then letting x = 1 and then x = -1, you get:

$(1+3)^{10} = \binom {10} 0 + \binom {10} 1 3^1 + \binom {10} 2 3^2 + \cdots + \binom {10} 9 3^9 + \binom {10} {10} 3^{10} \\ \\ (-1 +3)^{10 } = \binom {10} 0 - \binom {10} 1 3^1 + \binom {10} 2 3^2 - \cdots - \binom {10} 9 3^9 + \binom {10} {10} 3^{10}$

Adding the 2, the terms with opposing signs cancel out, leaving:

$4^{10} + 2^{10} = 2 \left ( \binom {10} 0 + \binom {10} 2 3^2 + \binom {10} 4 3^4 + \cdots \binom {10} {10} 3^{10} \right )$

Simplifying, you get the required result.

ps: my stoopid internet connection keeps dropping off! Hence taking so long to post above solution.

3. ## Re: Binomial Theorem HSC 1985 Q6

Originally Posted by boredofstudiesuser1
Could anyone help me with part ii of this question?

I finished part i and got 14. Im guessing part ii has something to do with using the 3+2x and maybe n=10 and some sort of sub for x. Part from that I'm lost.

Any help appreciated, thanks.

Screen_Shot_2018_06_17_at_4_18_55_pm.png

Screen_Shot_2018_06_17_at_4_19_04_pm.png
Pretty much bang on there. The other hint I would give is notice that you only see every second binomial coefficient. The first step to derive such a series is to substitute a value for x and then substitute the negative of that value.

4. ## Re: Binomial Theorem HSC 1985 Q6

Originally Posted by Drongoski
$(x + 3)^10 =$
Edit: Great, thank you so much.

5. ## Re: Binomial Theorem HSC 1985 Q6

Originally Posted by Trebla
Pretty much bang on there. The other hint I would give is notice that you only see every second binomial coefficient. The first step to derive such a series is to substitute a value for x and then substitute the negative of that value.
Thank you.

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