If $n\ge 2$, find $\binom 31 -\binom 42 +\binom 53....(-1)^{n-1} \binom {n+2}{n}$

Question

I have no clue on how to start solving. Can I get a hint to get me started?

HINT:its the coefficient of $x$ in $${(1+x)}^3-\frac{{(1+x)}^4}{x}+\frac{{(1+x)}^5}{x^2}...$$ recognise this as a GP ....... — Albus Dumbledore, Dec 28 '20 at 11:17

score 2 · Answer 1 · answered Dec 28 '20 at 11:18

2

Hint: This sum is equal to $$\frac{1}{2} \sum_{k=1}^n (-1)^{k-1} (k+1)(k+2)$$

answered Dec 28 '20 at 11:18

Asem Abdelraouf

1,260

And....? The hint is vague IMO – Shubham Johri Dec 28 '20 at 12:25
The OP is asking for a hint on how to start. There are many ways to go from here: one way would be to expand and use summation formulas. – Asem Abdelraouf Dec 28 '20 at 12:34

score 1 · Answer 2 · answered Dec 28 '20 at 11:19

1

Hint:

$${{r+2}\choose {r}} = {{r+2}\choose {2}} = \frac{(r+2)(r+1)}{2} $$ so the sum is $$\frac 12\sum_1^n (-1)^{r-1} (r+2)(r+1) \\ =\frac 12\sum_0^{\lfloor (n-1)/2 \rfloor}(2k+3)(2k+2) -\frac 12\sum_1^{\lfloor n/2 \rfloor} (2k+2)(2k+1)$$

answered Dec 28 '20 at 11:19

Vishu

14,469

How did you rearrange the terms in the last line? – Aditya Dec 28 '20 at 14:56
@Aditya I separated the odd and even terms, i.e. split the sum into $n=2k+1$ and $n=2k$. – Vishu Dec 28 '20 at 15:08
And why did you replace r with k? Also it’s not know where n is even or odd, so how did you assign values to the number of odd and even terms? – Aditya Jan 02 '21 at 08:35
@Aditya I have freedom to choose any variable index. The floors take care of that. The last term in the first sum will correspond to $r=2\lfloor (n-1)/2 \rfloor +1 = n$ if $n$ is odd and $n-1$ if $n$ is even. The last term in the second sum corresponds to $r=2\lfloor n/2 \rfloor = n$ if $n$ is even and $n-1$ if $n$ is odd. – Vishu Jan 02 '21 at 09:08
Why why replace it all? It seems that replacing $r \to 2k$ just makes it look more complicated – Aditya Jan 02 '21 at 09:31
@Aditya So then how do you plan on evaluating the $r$-sum? Do you have a simpler way? – Vishu Jan 02 '21 at 10:15
Never mind, I am just having a hard time understand what exactly is happening in the last line, why it changed to $\frac n2$ etc. I get that you split it into two series for when r is even and odd, but i don’t just don’t understand the reason for replace r. It’s just beyond me – Aditya Jan 02 '21 at 15:51
That’s exactly right. When r is even, I put $r=2k$ and when odd I put $r=2k+1$. If the summation notation confuses you, try writing out explicitly the terms to get a better understanding. – Vishu Jan 02 '21 at 17:05
Ok I got it. A few final questions though, shouldn’t $\frac {n-1}{2} + \frac n2 +1$ add up to $n$? – Aditya Jan 03 '21 at 14:17
@Aditya No, that’s $ n+\frac 12$. Why? – Vishu Jan 03 '21 at 14:22
There are $n$ terms so the total number of terms should add up to $n$ – Aditya Jan 03 '21 at 14:25
That’s not the total terms. You forgot the floors. Of the two floors, exactly one will return the argument unchanged, and the other returns the argument minus $\frac 12$. – Vishu Jan 03 '21 at 14:28

If $n\ge 2$, find $\binom 31 -\binom 42 +\binom 53....(-1)^{n-1} \binom {n+2}{n}$

2 Answers2