Binomial (1 Viewer)

[star*eyed]

Can anyone help with this question??

The expansion of (1 + ax)^n is given by 1-24x + 252x^2....... Find the values of a and n.

Thanx in advance

 

[zeek]

you are given the terms 1-24x+252x² and also the bionomial from which they were expanded... (1+ax)ⁿ

By examining the second term...

T₂=ⁿC₁(1)^n-1(ax)¹

but from the exapnsion we know the second term is -24x
.: -24x = ⁿC₁(1)^n-1(ax)¹ {ⁿC₁=n}
.: by eliminating we get -24=na [1]

also...

T₃=ⁿC₂(1)^n-2(ax)²

but T₃=252x²

.: 252x²=ⁿC₂(1)^n-2(ax)²

252x²= (n!/[2!(n-2)!]).1.a²x² {the factorial can be simplified into n(n-1)/2}

252=a²n(n-1)/2

.: 504=a²n² - a²n [2]

by substituting [1] in....

-72 = -a²n [3]

Now... by dividing [3] by [1] we get...
a=-3 also, by substituting a back into [1] we get n=8

Therefore, these terms were created by the expansion of the binomial (1-3x)⁸