Binomial Theorem HSC 1985 Q6 (1 Viewer)

[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.

 

[Drongoski]

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



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



Simplifying, you get the required result.




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

 

[Trebla]

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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View attachment 34571

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.

 

[boredofstudiesuser1]

Edit: Great, thank you so much.

 

[boredofstudiesuser1]

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.