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

tickboom

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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?

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Drdusk

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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

 

tickboom

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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!
 

CM_Tutor

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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



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