If the constant term in the binomial expansion of (x2-(1/x))n, n ∈ N is 15 then the value of n is equal to

Question

If the constant term in the binomial expansion of (x2-(1/x))n, n ∈ N is 15 then the value of n is equal to (A) 4 (B) 6 (C) 7 (D) 9

Tardigrade Admin · Accepted Answer

Tr+1 in (x2-(1/x))n is n Cr(x2)n-r(-1)r x-r = n Cr x2 n-3 r(-1)r Constant term = n Cr(-1)r if 2 n=3 r i.e. coefficient of x=0 hence n C2 n / 3(-1)2 n / 3=15= 6 C4 ⇒ n=6

Q. If the constant term in the binomial expansion of $(x^{2} - \frac{1}{x})^{n}, n \in N$ is 15 then the value of $n$ is equal to

4

6

7

9

$T_{r + 1}$ in $(x^{2} - \frac{1}{x})^{n}$ is
$^{n} C_{r} (x^{2})^{n - r} (- 1)^{r} x^{- r}$
$=^{n} C_{r} x^{2 n - 3 r} (- 1)^{r}$
Constant term $=^{n} C_{r} (- 1)^{r}$ if $2 n = 3 r$ i.e. coefficient of $x = 0$ hence $^{n} C_{2 n /3} (- 1)^{2 n /3} = 15 =^{6} C_{4} \Rightarrow n = 6$

Q. If the constant term in the binomial expansion of (x2−x1​)n,n∈N is 15 then the value of n is equal to

4

6

7

9

Tr+1​ in (x2−x1​)n is nCr​(x2)n−r(−1)rx−r =nCr​x2n−3r(−1)r Constant term =nCr​(−1)r if 2n=3r i.e. coefficient of x=0 hence nC2n/3​(−1)2n/3=15=6C4​⇒n=6

Q. If the constant term in the binomial expansion of $(x^{2} - \frac{1}{x})^{n}, n \in N$ is 15 then the value of $n$ is equal to

$T_{r + 1}$ in $(x^{2} - \frac{1}{x})^{n}$ is
$^{n} C_{r} (x^{2})^{n - r} (- 1)^{r} x^{- r}$
$=^{n} C_{r} x^{2 n - 3 r} (- 1)^{r}$
Constant term $=^{n} C_{r} (- 1)^{r}$ if $2 n = 3 r$ i.e. coefficient of $x = 0$ hence $^{n} C_{2 n /3} (- 1)^{2 n /3} = 15 =^{6} C_{4} \Rightarrow n = 6$