Question

Three consecutive terms in the expansion of (1+x)^n have coefficients 120,210, and 252. Find n.
BINOMIAL THEOREM QUESTION @amistre64

Answer 1

Should be a matter of solving the following:
\[\left(\begin{matrix}n \\ k\end{matrix}\right)=120 \\ \left(\begin{matrix}n \\ k+1\end{matrix}\right)=210 \\ \left(\begin{matrix}n \\ k+2\end{matrix}\right)=252\]

Answer 2

I dont know how to do matrices, is there another way of doing this?

Answer 3

I didn't any matrices...

Answer 4

\[\left(\begin{matrix}n \\ k\end{matrix}\right)=\frac{n!}{k!(n-k)!}\]

Answer 5

oh okay, got it let me try to solve it

Answer 6

okay, im lost now lol. how do i cancel the k's?

Answer 7

i know at some point n! reaches (n-k) factorial where then they both cancel out) but how do i get cancel out the other k!?

Answer 8

\[\frac{n!}{k!(n-k-2)! (n-k-1)(n-k)}=120 \\ \frac{n!}{k!(n-k-2)!(k+1)(n-k-1)}=210 \\ \frac{n!}{k!(n-k-2)!(k+1)(k+2)}=252\]

Answer 9

I don't think i've learned this yet. My teacher might go through this tomorrow.

Answer 10

yea, sorry didnt have to do this question, wasn't assigned

Answer 11

(n 0) (n 1) (n 2) ... (n k)

(n+1 0) (n 0) + (n 1) (n 1) + (n 2)

(n 0) + (n 1) + (n 1) + (n 2)

just athought

Answer 12

\[(n-k-1)(n-k)120=(k+1)(n-k-1)210=(k+1)(k+2)252 \\ \]
\[(n-k)120=(k+1)210 \\ (n-k-1)210=(k+2)252 \]

Answer 13

\[(n-k)210-210=252k+504 \\ \frac{(k+1)(210)}{120}(210)-210=252k+504\]

Answer 14

so you can solve for k

Answer 15

and then you could go back and solve for n

Answer 16

okay, awsome thanks

Answer 17

how about we simply let n=10 ...

Answer 18

yea that what i got from trial and error

Answer 19

I got n=9 by doing Freckles way

Answer 20

k=3 \[(n-3)120=(3+1)210 \\n-3=\frac{(3+1)210}{120}=\frac{4(21)}{12}=\frac{21}{3}=7 \\ n-3=7 \]

Answer 21

What did you get for k @M0j0jojo ?

Answer 22

i got 3 aswell maybe i probably messed on calculating n, let me check again

Answer 23

yea found my mistake. got n=10

Answer 24

Also there is a lot of mistakes we can make in this problem since there so many steps.

Answer 25

I like @amistre64 way of saying let's just make n=10.

Answer 26

lol, it just seemed easier to me that way

Answer 27

lol, that what i did cuz from memory i know 10c5=252 so i was like oh maybe n=10

Answer 28

if the 10 row was too large ... i was going to say 8 and just try and error it into submission

Answer 29

\[n!=\int_{0}^{\infty} x^ne^{-x}~dx\]

wonder if this is useful

Answer 30

o.o

Answer 31

\[\left(\begin{matrix}n \\ k+1\end{matrix}\right)=\frac{n-k}{k+1}\left(\begin{matrix}n \\ k\end{matrix}\right) \\ \left(\begin{matrix}n \\ k+2\end{matrix}\right)=\frac{n-k-1}{k+2} \left(\begin{matrix}n \\ k+1\end{matrix}\right)\]

Answer 32

These probably would have gave us better but not much better equations to solve

Answer 33

I'm interesting in the integral way if you know how to approach that @amistre64

Answer 34

rolling it around, not sure if its useful tho :) i cant recall the gamma function for nuthin off the top of my head

Answer 35

120 C 1 = 120
120 C 2 not= 210

nC2 = 120

120 = x(x-1)/2
0 = n^2 -n-240; x=16

16C2 = 120
16C3 = 560 not it


nC3 = 120
n(n-1)(n-2) = 720
n(n-1)(n-2) - 720 = 0

n = 10

10C3 = 120
10C4 = 210
10C5 .... would prolly have been my ultimate approach

Answer 36

It would of been easier with just listing out pasca'ls triangle