binomial expansion

Question

if  (2r + 3 )th  and (r-1) th terms in the expansion of (1+ x) ¹⁵ have equal coefficiants  then r equals 

  

• 3
• 4
• 5
• 6

 

Answer 1

The coefficients for that expansion are:

 

₁₅C_{0 15}C_{1 15}C₂ ... ₁₅C_{13 15}C_{14 15}C₁₅

 

or:

 

1 15 105 ... 105 15 1

 

 

 

The row is symmetric. So, the 1st and 16th terms have the same coefficients, as do terms 2 and 15, 3 and 14, 4 and 13, 5 and 12, 6 and 11, 7 and 10, and 8 and 9.

 

Notice that, in each case, the indices of the pair of terms add up to 17: 1 + 16 = 2 + 15 = ... = 8 + 9 = 17.

 

 

 

So, whatever our terms are, (2r + 3) + (r - 1) = 17. Solve for r:

 

2r + 3 + r - 1 = 17

3r + 2 = 17

3r = 15

r = 5

 

 

 

To check our work, r = 5 means our terms are:

 

2r + 3 = 2(5) + 3 = 10 + 3 = 13

 

and

 

r - 1 = 5 - 1 = 4

 

The 13th term has a coefficient of ₁₅C₁₂ = 455.

The 4th term has a coefficient of ₁₅C₃ = 455.

 

That checks out!