2
$\begingroup$

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 $

  • 0
    Sorry, what is $C_i$ ? Is it $\binom ni$?2010-12-11
  • 0
    Andres Caicedo: Fixed.2010-12-11
  • 0
    This looks like it will be a dupe. I am unable to find it though. Anyone?2010-12-11

4 Answers 4

1

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.

  • 0
    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.2010-12-11
  • 0
    @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".2010-12-11
  • 0
    @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...2010-12-11
  • 0
    @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.2010-12-11
  • 0
    @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.2010-12-11
  • 0
    @Moron: Perhaps you read too much into the *name* "calculus free" - which I used merely to be consistent with terminology in other posts here. My point was to illustrate what the integration amounts to from a generating function perspective. Your proof can be derived completely *mechanically* from this equivalence.2010-12-11
6

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

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

  • 0
    Nice: +12010-12-11
  • 0
    +1...Nice one Moron,but it might be weird that I feel the calculus method is a bit more easy to understand ;-)2010-12-11
  • 0
    @Deb: Calculus approach is more natural to me too and is in fact much more general. I don't see any surprises there :-)2010-12-11
  • 1
    Oh, yes! Now I even remember where I saw this first: In "Problem solving through problems", by Loren C. Larson.2010-12-11
5

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

  • 0
    Thanks a lot I got $\frac{3^{n+1}-1}{n+1}$2010-12-11
  • 0
    @Debanjan: That's correct now.2010-12-11
  • 0
    Opps earlier it was a typo.Btw your hints always just made my day:)2010-12-11
3

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.

  • 0
    @Deb: You recently claimed you didn't know calculus...2010-12-11
  • 0
    What I said is that I haven't started off with calculus ;-)2010-12-11