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I tried to using $2^{999} = \sum^{999}_{i=1}\binom{999}{i}$ but I just don't know how to continue from here.

Tienanh Nguyen
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1 Answers1

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$$\sum^{n}_{i=1}\binom{n}{i}x^i=(1+x)^{n}$$

Derive once:

$$\sum^{n}_{i=1}\binom{n}{i}i x^{i-1}=n(1+x)^{n-1}$$

Derive again:

$$\sum^{n}_{i=1}\binom{n}{i}i(i-1) x^{i-2}=n(n-1)(1+x)^{n-2}$$ $$\sum^{n}_{i=1}\binom{n}{i}i^2x^{i-2}=\sum^{n}_{i=1}\binom{n}{i}ix^{i-2}+n(n-1)(1+x)^{n-2}$$

Take $x=1$ combine both results to obtain what you want.

rlartiga
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  • I'm really sorry but I mistake 2^999 + 999000 with 2^999 * 999000. In the case that your answer is still correct, after replace x = 1 you get on the left-side $\sum^{n}_{i=1}\binom{n}{i}i+n(n-1)(2)^{n-2}$. What do you do from here ? – Tienanh Nguyen Jan 25 '17 at 22:16
  • I'm sorry. I really don't want to nag you the answer, but I've been spending 2 hours trying to work this out but all I got is n(n+1)2^(n-2) by replacing the function when you first derive the original one to the second one and it is really frustrating. Could you maybe tell me what I'm doing wrong? – Tienanh Nguyen Jan 26 '17 at 01:17
  • @TienanhNguyen: all you got is the answer, just replace $n$ with $999$. But have a look at the combinatorial argument given here (http://math.stackexchange.com/a/1923224/44121), too. – Jack D'Aurizio Jan 26 '17 at 01:59
  • but doesn't that mean the RHS for my question should be 2^997*999000. Does that mean my question is wrong. It is homework. – Tienanh Nguyen Jan 26 '17 at 02:13
  • yes I verify. It is wrong. Thank you guys for helping – Tienanh Nguyen Jan 26 '17 at 03:15
  • @TienanhNguyen I'm sorry for replying too late. I hope it's everything ok. – rlartiga Jan 26 '17 at 12:16
  • Yes, I figured out the answer and the correct question. No worry anymore! – Tienanh Nguyen Jan 26 '17 at 12:17