The term independent of x in the expansion of (x+(1/x2))6 is

Question

The term independent of x in the expansion of (x+(1/x2))6 is (A) 20 (B) 15 (C) 6 (D) 1

Tardigrade Admin · Accepted Answer

The general term of (x+(1/x2))6 is Tr+1= 6 Cr xr((1/x2))6-r = 6 Cr xr-12+2 r For independent of x, r-12+2 r=0 ⇒ 3 r=12 ⇒ r=4 ∴ Required term = 6 C4=(6 × 5/2 × 1)=15

Q. The term independent of $x$ in the expansion of $(x + \frac{1}{x ^{2}})^{6}$ is

20

15

6

1

0

The general term of $(x + \frac{1}{x ^{2}})^{6}$ is
$T_{r + 1} =^{6} C_{r} x^{r} (\frac{1}{x ^{2}})^{6 - r}$
$=^{6} C_{r} x^{r - 12 + 2 r}$
For independent of $x$ ,
$r - 12 + 2 r = 0$
$\Rightarrow 3 r = 12$
$\Rightarrow r = 4$
$∴$ Required term $=^{6} C_{4} = \frac{6 \times 5}{2 \times 1} = 15$

Q. The term independent of x in the expansion of (x+x21​)6 is

20

15

6

1

0

The general term of (x+x21​)6 is Tr+1​=6Cr​xr(x21​)6−r =6Cr​xr−12+2r For independent of x, r−12+2r=0 ⇒3r=12 ⇒r=4 ∴ Required term =6C4​=2×16×5​=15

Q. The term independent of $x$ in the expansion of $(x + \frac{1}{x ^{2}})^{6}$ is

The general term of $(x + \frac{1}{x ^{2}})^{6}$ is
$T_{r + 1} =^{6} C_{r} x^{r} (\frac{1}{x ^{2}})^{6 - r}$
$=^{6} C_{r} x^{r - 12 + 2 r}$
For independent of $x$ ,
$r - 12 + 2 r = 0$
$\Rightarrow 3 r = 12$
$\Rightarrow r = 4$
$∴$ Required term $=^{6} C_{4} = \frac{6 \times 5}{2 \times 1} = 15$