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How to sum up this series :

$$2C_o + \frac{2^2}{2}C_1 + \frac{2^3}{3}C_2 + \cdots + \frac{2^{n+1}}{n+1}C_n$$

Any hint that will lead me to the correct solution will be highly appreciated.

EDIT: Here $C_i = ^nC_i $

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Sorry, what is $C_i$ ? Is it $\binom ni$? – Andrés E. Caicedo Dec 11 '10 at 16:18
Andres Caicedo: Fixed. – Quixotic Dec 11 '10 at 16:30
This looks like it will be a dupe. I am unable to find it though. Anyone? – Aryabhata Dec 11 '10 at 17:42
up vote 5 down vote accepted

Hint: Use binomial expansion for $(1+x)^n$ and integrate once. Then choose an appropriate value for $x$.

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Thanks a lot I got $\frac{3^{n+1}-1}{n+1}$ – Quixotic Dec 11 '10 at 16:25
@Debanjan: That's correct now. – Timothy Wagner Dec 11 '10 at 16:26
Opps earlier it was a typo.Btw your hints always just made my day:) – Quixotic Dec 11 '10 at 16:27

For an elementary method which does not use calculus:

Notice that $\displaystyle \dfrac{{n \choose k}}{k+1} = \dfrac{1}{n+1} {n+1 \choose k+1}$

Thus your sum is

$$\sum_{k=0}^{n} \dfrac{1}{n+1} {n+1 \choose k+1} 2^{k+1} = \dfrac{\sum_{k=0}^{n+1} {n+1 \choose k} 2^k -1}{n+1} = \dfrac{3^{n+1} - 1}{n+1}$$

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Nice: +1 – Timothy Wagner Dec 11 '10 at 17:53
+1...Nice one Moron,but it might be weird that I feel the calculus method is a bit more easy to understand ;-) – Quixotic Dec 11 '10 at 18:02
@Deb: Calculus approach is more natural to me too and is in fact much more general. I don't see any surprises there :-) – Aryabhata Dec 11 '10 at 18:04
Oh, yes! Now I even remember where I saw this first: In "Problem solving through problems", by Loren C. Larson. – Andrés E. Caicedo Dec 11 '10 at 19:07

Let's assume $C_i=\binom ni$. I'll give a solution that is not precalculus level. Consider first the equality $$ (1+x)^n=C_0+xC_1+x^2C_2+\dots+x^nC_n. $$ This is the binomial theorem.

Integrate from 0 to t. On the left hand side we get $\frac{(1+t)^{n+1}-1}{n+1}$ and on the right hand side $\sum \frac1{i+1}t^{i+1}C_i$.

Now set $t=2$, and a bit of algebra gives you the answer you want.

Pretty sure there is an elementary approach as well.

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+1 Yes,that what I did :) – Quixotic Dec 11 '10 at 16:26
@Deb: You recently claimed you didn't know calculus... – Aryabhata Dec 11 '10 at 17:42
What I said is that I haven't started off with calculus ;-) – Quixotic Dec 11 '10 at 17:51

REMARK $\ $ The various approaches are all equivalent. Namely, suppose that we desire to prove without calculus the identity arising from integrating the binomial formula, viz. $$\rm (1 + x)^{n+1}\ =\ 1 + \sum_{k=1}^{n+1}\: \frac{n+1}{k+1} {n\choose k}\ x^{k+1}$$

Comparing coefficients reduces it to the identity

$$\rm \quad\quad\ {n+1 \choose k+1}\ =\ \frac{n+1}{k+1} {n\choose k} $$

which is precisely the identity employed in Moron's "calculus free" approach.

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FWIW, the way I came up with that identity is by using $\displaystyle {n \choose k} = \dfrac{n!}{(n-k)!k!}$. So claiming that there are equivalent is a bit of a stretch, especially since no knowledge of calculus is required in the proof of that identity. – Aryabhata Dec 11 '10 at 20:12
@Moron: The point is that the two proofs are equivalent from the generating function perspective. This is a mathematical fact that is independent of how you "came up with that identity". – Bill Dubuque Dec 11 '10 at 20:19
@Bill: I was mainly responding to the "calculus free" comment which kind of implies that it is not possible without calculus. Of course, new techniques might be discovered which render two previously different approaches as "equivalent" (so I agree with you), but they could still be useful for the different insights they provide, historical reasons etc. Anyway... – Aryabhata Dec 11 '10 at 20:28
@Moron: It's still not clear to me what your point is, but perhaps you have read more into my post than what was intended. – Bill Dubuque Dec 11 '10 at 20:37
@Bill: The point is that the definition of "equivalent" is subjective. For instance, from one point of view any two tautologies are equivalent. My main objection is putting calculus free in quotes, implying that calculus is required, while it is not. Anyway, I will stop bothering you now. – Aryabhata Dec 11 '10 at 20:41

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