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

#### 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:

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

For the LHS we get

Equating the coefficients to the x^r term of the RHS which is we get the solution

• B1andB2 and tickboom

#### tickboom

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

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

For the LHS we get

Equating the coefficients to the x^r term of the RHS which is we get the solution

Legendary. Thank you!

#### YonOra

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

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

For the LHS we get

Equating the coefficients to the x^r term of the RHS which is we get the solution

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

and then select the term in , which is:

The term in on the RHS is

and so equating the coefficients of these terms gives

as required

• Drdusk and nunyabeezwax