Total number of term, that are dependent on the value of x, in the expansion of (x2-2+(1/x2))n is equal to

Question

Total number of term, that are dependent on the value of x, in the expansion of (x2-2+(1/x2))n is equal to (A) 2 n+1 (B) 2 n (C) n (D) n+1

Tardigrade Admin · Accepted Answer

⇒ ((x2-1)2 n/x2 n) T r +1=( 2 n C r ⋅( x 2)2 n - r (-1) r / x 2 n ) At r = n, term is independent Hence, total number of term =2 n+1-1=2 n independent of x

Q. Total number of term, that are dependent on the value of $x$ , in the expansion of $(x^{2} - 2 + \frac{1}{x ^{2}})^{n}$ is equal to

$2 n + 1$

$2 n$

$n$

$n + 1$

$\Rightarrow \frac{( x ^{2} - 1 ) ^{2 n}}{x ^{2 n}}$
$T_{r + 1} = \frac{^{2 n} C _{r} \cdot ( x ^{2} ) ^{2 n - r} ( - 1 ) ^{r}}{x ^{2 n}}$
At $r = n$ , term is independent Hence, total number of term $= 2 n + 1 - 1 = 2 n$ independent of $x$

Q. Total number of term, that are dependent on the value of x, in the expansion of (x2−2+x21​)n is equal to

2n+1

2n

n

n+1

⇒x2n(x2−1)2n​ Tr+1​=x2n2nCr​⋅(x2)2n−r(−1)r​ At r=n, term is independent Hence, total number of term =2n+1−1=2n independent of x

Q. Total number of term, that are dependent on the value of $x$ , in the expansion of $(x^{2} - 2 + \frac{1}{x ^{2}})^{n}$ is equal to

$2 n + 1$

$2 n$

$n$

$n + 1$

$\Rightarrow \frac{( x ^{2} - 1 ) ^{2 n}}{x ^{2 n}}$
$T_{r + 1} = \frac{^{2 n} C _{r} \cdot ( x ^{2} ) ^{2 n - r} ( - 1 ) ^{r}}{x ^{2 n}}$
At $r = n$ , term is independent Hence, total number of term $= 2 n + 1 - 1 = 2 n$ independent of $x$