Binomial theorem Q (1 Viewer)

[shaon0]

By considering (1+x)^n where t{k} is a the coefficient of x^k. Find the value for which t{1}, t{2}, t{3} are in arithmetic progression. '{}' means subscript.

 

[Timothy.Siu]

t1 = nC1
t2=nC2
t3=nC3

for arithmetic progression, nC2-nC1=nC3-nC2
2(nC2)=nC3+nC1
2(n!/[2!(n-2)!]=n!/3!(n-3)! + n!/(n-1)!
simplifying..
n(n-1)=n(n-1)(n-2)/6 + n
since n=/=0
n-1=(n-1)(n-2)/6+1
multiplying by 6
6n-6=n²-3n+2+6
n²-9n+14=0
(n-7)(n-2)=0

n=7 or n=2 but n>2
so, n=7

 

[shaon0]

t1 = nC1
t2=nC2
t3=nC3

for arithmetic progression, nC2-nC1=nC3-nC2
2(nC2)=nC3+nC1
2(n!/[2!(n-2)!]=n!/3!(n-3)! + n!/(n-1)!
simplifying..
n(n-1)=n(n-1)(n-2)/6 + n
since n=/=0
n-1=(n-1)(n-2)/6+1
multiplying by 6
6n-6=n²-3n+2+6
n²-9n+14=0
(n-7)(n-2)=0

n=7 or n=2 but n>2
so, n=7

Thnaks for the help.

 

[lyounamu]

you are pro, timmy. teach me maths at cosmolgy camp xD.