# Alternative way to prove nCr + 2nCr-1 + nCr-2 = n+2Cr (1 Viewer)

#### tickboom

##### Member
Hi all, I'm a bit stuck on part ii) of this question. Part i) was easy enough, and I also know how to prove part ii) using Pascal's identity. But I can't see the link between part i) and part ii), which seems to be implied by the word "hence" in the question. Any ideas?

#### Drdusk

##### π
Moderator
Hi all, I'm a bit stuck on part ii) of this question. Part i) was easy enough, and I also know how to prove part ii) using Pascal's identity. But I can't see the link between part i) and part ii), which seems to be implied by the word "hence" in the question. Any ideas?

View attachment 29657
From (i) expand the x^n and (1+ 1/x)^n into each other:

$\bg_white \text{LHS} = (1+x)^n(1+2x + x^2)$

Equate the x^r terms from the RHS and the LHS.

For the LHS we get

$\bg_white {n \choose r}x^r \times 1$

$\bg_white +$

$\bg_white {n \choose r-1}x^{r-1} \times 2x$

$\bg_white +$

$\bg_white {n \choose r-2}x^{r-2} \times x^2$

Equating the coefficients to the x^r term of the RHS which is $\bg_white {n+2 \choose r}x^{r}$ we get the solution

$\bg_white {n \choose r}+ 2{n \choose r-1}+ {n \choose r-2}= {n+2 \choose r}$

#### tickboom

##### Member
From (i) expand the x^n and (1+ 1/x)^n into each other:

$\bg_white \text{LHS} = (1+x)^n(1+2x + x^2)$

Equate the x^r terms from the RHS and the LHS.

For the LHS we get

$\bg_white {n \choose r}x^r \times 1$

$\bg_white +$

$\bg_white {n \choose r-1}x^{r-1} \times 2x$

$\bg_white +$

$\bg_white {n \choose r-2}x^{r-2} \times x^2$

Equating the coefficients to the x^r term of the RHS which is $\bg_white {n+2 \choose r}x^{r}$ we get the solution

$\bg_white {n \choose r}+ 2{n \choose r-1}+ {n \choose r-2}= {n+2 \choose r}$
Legendary. Thank you!

#### YonOra

##### Well-Known Member
From (i) expand the x^n and (1+ 1/x)^n into each other:

$\bg_white \text{LHS} = (1+x)^n(1+2x + x^2)$

Equate the x^r terms from the RHS and the LHS.

For the LHS we get

$\bg_white {n \choose r}x^r \times 1$

$\bg_white +$

$\bg_white {n \choose r-1}x^{r-1} \times 2x$

$\bg_white +$

$\bg_white {n \choose r-2}x^{r-2} \times x^2$

Equating the coefficients to the x^r term of the RHS which is $\bg_white {n+2 \choose r}x^{r}$ we get the solution

$\bg_white {n \choose r}+ 2{n \choose r-1}+ {n \choose r-2}= {n+2 \choose r}$
No video recording of your working? Kinda disappointing...

#### CM_Tutor

##### Well-Known Member
Hi all, I'm a bit stuck on part ii) of this question. Part i) was easy enough, and I also know how to prove part ii) using Pascal's identity. But I can't see the link between part i) and part ii), which seems to be implied by the word "hence" in the question. Any ideas?

View attachment 29657
You could also expand

$\bg_white \text{LHS } = \left(1+\frac{1}{x}\right)^n\left(x^n+2x^{n+1}+x^{n+2}\right)$

and then select the term in $\bg_white x^r$, which is:

\bg_white \begin{align*} \text{Required term in x^r } &= \binom{n}{n-r}\left(\frac{1}{x}\right)^{n-r} \times x^n + \binom{n}{n-r+1}\left(\frac{1}{x}\right)^{n-r+1} \times 2x^{n+1} + \binom{n}{n-r+2}\left(\frac{1}{x}\right)^{n-r+2} \times x^{n+2} \\ &= \binom{n}{n-r} \times x^{n-(n-r)} + \binom{n}{n-r+1} \times 2x^{n+1-(n-r+1)} + \binom{n}{n-r+2} \times x^{n+2-(n-r+2)} \\ &= \left[\binom{n}{n-r} + 2\binom{n}{n-r+1} + \binom{n}{n-r+2}\right] x^r \\ &= \left[\binom{n}{n-(n-r)} + 2\binom{n}{n-(n-r+1)} + \binom{n}{n-(n-r+2)}\right] x^r \text{ using the symmetry property \binom{n}{k} = \binom{n}{n-k}} \\ &= \left[\binom{n}{r} + 2\binom{n}{r-1} + \binom{n}{r-2}\right] x^r \end{align*}

The term in $\bg_white x^r$ on the RHS is

$\bg_white \binom{n+2}{r}x^r$

and so equating the coefficients of these terms gives

$\bg_white \binom{n}{r} + 2\binom{n}{r-1} + \binom{n}{r-2} = \binom{n+2}{r}$

as required