Question

Determine the value of n in the expansion of (a+b)^n in descending powers of a, it is found that two adjacent coefficients are each equal to 126.

Answer 1

General term is \[nCr(a)^{n-r}(b)^{r}\]

Answer 2

Construct a Pascal's triangle.

Answer 3

I let term 1= the first 126
r+1=1
r=0
and then plugged r=0 into the r in the general term\[126=nC0(a)^{n-0}(b)^{0}\]
when simplified i get 126=a^n

Answer 4

yea i did that i got n=9, but i wanted know if there was a faster way

Answer 5

\[
^nC_r = ^nC_{r+1} = 126
\]

Answer 6

right

Answer 7

\[\frac{ n! }{ (n-r)!r! }=\frac{ n! }{ (n-r+1)!(r+1)! }=126\]

Answer 8

now how do i go about solving this abomination?

Answer 9

Divide the first by the second and equate it to 1. Find r in terms of n.
Then put r in the first one equate it to 126 and solve for n.

Answer 10

What do you mean by the first and the second? first=nCr? second= nCr+1?

Answer 11

yes.

Answer 12

\[
\frac{ n! }{ (n-r)!r! } \div \frac{ n! }{ (n-r+1)!(r+1)! }=1
\]

Answer 13

okay

Answer 14

\[
\frac{ n! }{ (n-r)!r! } \div \frac{ n! }{ (n-r+1)!(r+1)! }=1 \\
\frac{ \cancel{n!} }{ (n-r)!r! } \times \frac{ (n-r+1)!(r+1)! }{\cancel{n!}} = 1 \\

\]

Answer 15

yea that what i got, now i dont know what to do

Answer 16

\[\frac{ n-r+1)!(r+1)! }{ (n-r)!r!}=1\]

Answer 17

\[
\frac{ \cancel{n!} }{ (n-r)!r! } \times \frac{ (n-r+1)!(r+1)! }{\cancel{n!}} = 1 \\
\frac{ (n-r+1)*(n-r)! *(r+1)*r! }{(n-r)!r!} = 1 \\


\]

Answer 18

No. The above is based on the previous result which was wrong.

Answer 19

wait how did you (n-r)! on top

Answer 20

\[
^nC_r = ^nC_{r+1} = 126 \\
\frac{n!}{(n-r)!*r!} = \frac{n!}{(n-r-1)!*(r+1)!} = 126 \\
\frac{n!}{(n-r)!*r!} \div \frac{n!}{(n-r-1)!*(r+1)!} = 1 \\
\frac{n!}{(n-r)!*r!} \times \frac{(n-r-1)!*(r+1)!}{n!} = 1 \\
\frac{(n-r-1)!*(r+1)!}{(n-r)!*r!} = 1 \\
\frac{(n-r-1)!*(r+1) * r!}{(n-r) * (n-r-1)!*r!} = 1 \\
\frac{r+1}{n-r} = 1 \\
r + 1 = n - r \\
2r = n - 1 \\
r = \frac 12 (n-1)

\]

Answer 21

I think Pascal's triangle is the easier method.

Answer 22

now that i see that, I think the same lol

Answer 23

Thanks for the help, Pascal's triangle is the way to go for this it seems.

Answer 24

You are welcome.