The first three terms in the expansion of (1+ax)n(n≠ 0) are 1,6x and 16x2 . Then, the value of a and n are respectively

Question

The first three terms in the expansion of (1+ax)n(n≠ 0) are 1,6x and 16x2 . Then, the value of a and n are respectively (A) 2 and 9 (B) 3 and 2 (C) 2/3 and 9 (D) 3/2 and 6

Tardigrade Admin · Accepted Answer

Given, T1=nC0=1 ?.(i) T2=nC1ax=6x ⇒ nax=6x ...(ii) and T3=nC2(ax)2=16x2 ⇒ (n(n-1)/2)a2x2=16x2 ?.(iii) On solving Eqs. (ii) and (iii), we get a=(2/3) and n=9

Q. The first three terms in the expansion of $(1 + a x)^{n} (n \neq = 0)$ are $1, 6 x$ and $16 x^{2}$ . Then, the value of a and n are respectively

2 and 9

3 and 2

2/3 and 9

3/2 and 6

Given, $T_{1} =^{n} C_{0} = 1$ ?.(i) $T_{2} =^{n} C_{1} a x = 6 x$ $\Rightarrow$ $na x = 6 x$ ...(ii) and $T_{3} =^{n} C_{2} (a x)^{2} = 16 x^{2}$ $\Rightarrow$ $\frac{n ( n - 1 )}{2} a^{2} x^{2} = 16 x^{2}$ ?.(iii) On solving Eqs. (ii) and (iii), we get $a = \frac{2}{3}$ and $n = 9$

Q. The first three terms in the expansion of (1+ax)n(n=0) are 1,6x and 16x2 . Then, the value of a and n are respectively