Question

If ${\left(1+ax\right)}^n=1+8x+24x^2+........$ then

• A. $a=3$
• B. $n=5$
• C. $a=2$
• D. $n=4$

Answer 1

Answer: (C), (D)

The correct options are
C $a=2$
D $n=4$

$(1+ax{)}^n=1+8x+24x^2+.......$

but

$(1+ax{)}^n={}^n{C}_0+{}^n{C}_1(ax)+{}^n{C}_2(ax{)}^2+......$

$=1+nax+\left(\frac{n(n-1)}{2}\right)(ax{)}^2+........$

by comparing

$\therefore na=8$

and

$\left(\frac{n(n-1)}{2}\right)a^2=24$

$\left(\frac{na(na-a)}{2}\right)=24$

$8(8-a)=48$

$8-a=6$

$\therefore a=2$

and

$na=8$

$\therefore n=\left(\frac{8}{a}\right)=4$

$\therefore n=4$