Binomial theorem help (1 Viewer)

[luvotomy]

The coefficients of x^4 and x^5 in the expansion of (3-x)^n are equal in magnitude but opposite in sign. find the value of n.

ans: n=19

thankss

 

[lyounamu]

The coefficients of x^4 and x^5 in the expansion of (3-x)^n are equal in magnitude but opposite in sign. find the value of n.

ans: n=19

thankss

x^4: nC4 . (3)^(n-4) . (-x)^4 = nC4 . 3^(n-4) x^4
x^5: nC5 . (3)^(n-5) . (-x)^5 = -nC5 . 3^ (n-4) x^5

So nC4 . 3^(n-4) = nC5 . (3)^(n-5) since they are same in magnitude (ignoring the sign)


3nC4 = nC5
3 . n!/(n-4)!4! = n!/(n-5)5!
15/(n-4)! = 1/(n-5)!
15/(n-4)(n-5)! = 1/(n-5)!
15/(n-4) = 1
15 = n-4
n= 19

Hope my working out explained itself.

 

[tommykins]

回复: Binomial theorem help

I'll have a shot, but binomial is my weakest 3u topic.

(3-x)^n = nck.(-x)^k.3^n-k

When k = 4 and 5, the coeff's are equal in magnitude.

nc4.3^n-4 = nc5.3^n-5
nc4.3^n.3^-4 = nc5.3^n.3^-4.3^-1.3^n
nc4 = nc5.3^-1
3nc4 = nc5

Use the formula and solve from there.