$1\binom{20}1+2\binom{20}2+3\binom{20}3+\dots+19\binom{20}{19}+20\binom{20}{20}$

Question

$$1\binom{20}1+2\binom{20}2+3\binom{20}3+\dots+19\binom{20}{19}+20\binom{20}{20}$$

I solved it by letting the sum be $S$, then adding the sum to itself but taking the terms from last to first and then using the property $\binom{n}r=\binom{n}{n-r}$ to take the combination part common and simply use the summation of $\binom{n}r$ from $r=1$ to $n$ (which is equal to $2^n-1$) to get the sum.

However, I tried to do it in a different way but was unable to do that.

What I intended to do was consider the binomial coefficients to be that of the terms of expansion $(x+y)^{20}$ and the numbers $1$, $2$, $3$, up to $20$ to be in the value of the expansion if we substitute $x$ and $y$ with two specific values. Then I can just calculate the value of the expansion for the values and subtract $\binom{20}0x^{20}$ to get the required sum. However that clearly doesn't seem to work as:

Let $$S=(x+y)^{20}$$ $$\implies S=\binom{20}0x^{20}+\binom{20}1x^{19}y+\binom{20}2x^{18}y^2+\dots+\binom{20}{20}y^{20}$$

Now let $$x^{19}y=1$$ and $$x^{18}y^2=2$$

Dividing the equations, we get $$\frac xy= \frac12$$

Now again let $$x^{18}y^2=2$$ and $$x^{17}y^3=3$$ Dividing the equations, we get $$\frac xy= \frac23$$

So obviously nothing makes sense if I do it this way since I have already deduced that $\frac xy = \frac 12$ previously.

Is this approach of mine nonsensical or am I proceeding incorrectly? If yes, why?

@lulu Thanks for the comment. I have actually done it by differentiation too and I know how to proceed with that. My question is about the issue with my way of wanting to approach the problem. — AltercatingCurrent, Apr 01 '23 at 18:33
Your approach won't work because, as you've noted, the equations you want to solve simultaneously aren't simultaneously solvable. — J.G., Apr 01 '23 at 18:34
@AltercatingCurrent But I can not understand your approach. With the derivative, one just takes $x=1$ so those terms cease to appear. Of course, you could reverse engineer something...we know that the sum you want is associated to the function $20(1+x)^{19}$ so you could simply start with that. Seems artificial, though. — lulu, Apr 01 '23 at 18:35
Yeah, there is no one be one pair $x,y$ such that $x^ky^{20-k}=k$ for $k=1,\dots,20.$ That approach won't work. That's $20$ equations in $2$ variables, which makes it highly unlikely to work. — Thomas Andrews, Apr 01 '23 at 18:39
@J.G. how do I determine if it is solvable or not? Do I have to trivially check for it? — AltercatingCurrent, Apr 01 '23 at 18:41
@AltercatingCurrent It can't work that way. The terms $x^i y^{20-i}$ are in a geometric progression, so you'll never get them to fit the arithmetic progression $1,2,3,\dots$ — dxiv, Apr 01 '23 at 18:41
@ThomasAndrews how do I determine how "likely" is it solvable? — AltercatingCurrent, Apr 01 '23 at 18:42
You've proven it's not by showing it would give multiple values to $x/y$. (One subtlety: $y$ cannot be $0$, so there can't be multiple $k$ satisfying $x=ky$.) — J.G., Apr 01 '23 at 18:43
@lulu alright. I actually was just curious about why could I actually not solve for those two variables. dxiv's comment gave me a pretty straight reasoning for why it cannot so I think I understand now. — AltercatingCurrent, Apr 01 '23 at 18:45
@AltercatingCurrent My statement isn't logical, just an expectation. In general, $n$ equations in $n$ variables usually determines a finite set of points. — Thomas Andrews, Apr 01 '23 at 18:56
It's not impossible that $1,000,000$ equations in two variables share a common solution, but they have to be tightly related equations in a way that makes it something you would be surprised to find in general. — Thomas Andrews, Apr 01 '23 at 19:02
Does this answer your question? Find the value of $\sum_{k=1}^{n} k \binom {n} {k}$ — Anne Bauval, Apr 01 '23 at 20:02
Does this answer your question? Sum of $k {n \choose k}$ is $n2^{n-1}$ — Macavity, Apr 02 '23 at 07:02

score 0 · Answer 1 · answered Apr 01 '23 at 19:06

0

Here's an alternative approach that uses the binomial theorem but does not require differentiation: $$\sum_{k=1}^n k \binom{n}{k} = n \sum_{k=1}^n \binom{n-1}{k-1} = n \sum_{k=0}^{n-1} \binom{n-1}{k} = n 2^{n-1} $$

answered Apr 01 '23 at 19:06

RobPratt

45,619

It was a (multi)duplicate. – Anne Bauval Apr 01 '23 at 20:04
I proposed a duplicate in the comments section. Here is another one, which is an exact duplicate of your answer: https://math.stackexchange.com/a/683740/386889 – Anne Bauval Apr 01 '23 at 20:55

$1\binom{20}1+2\binom{20}2+3\binom{20}3+\dots+19\binom{20}{19}+20\binom{20}{20}$

1 Answers1