If the sum of the coefficients of x^7 and x^14 in the expansion of (1x^3-x^4)^n,x 0, is ze…

Question

If the sum of the coefficients of ${x}^{7}$ and ${x}^{14}$ in the expansion of ${(\frac{1}{{x}^{3}}-{x}^{4})}^{n},x
e 0$, is zero, then the value of $n$ is $____$ .

Answer: 21

Accepted Answer

For

${(\frac{1}{{x}^{3}}-{x}^{4})}^{n}={({x}^{-3}-{x}^{4})}^{n},$

let us write the general term in the expansion.

The general term is

${T}_{r+1}=(\frac{n}{r})({x}^{-3}{)}^{\, n-r}(-{x}^{4}{)}^{r}$

$=(\frac{n}{r})(-1{)}^{r}{x}^{-3(n-r)}{x}^{4r}$

$=(\frac{n}{r})(-1{)}^{r}{x}^{-3n+3r+4r}$

$=(\frac{n}{r})(-1{)}^{r}{x}^{-3n+7r}.$

So, the coefficient of ${x}^{k}$ comes from the term for which

$-3n+7r=k.$

Now find the coefficients of ${x}^{7}$ and ${x}^{14}$.

For ${x}^{7}$:

$-3n+7r=7$

$7r=3n+7$

$r=\frac{3n+7}{7}.$

For ${x}^{14}$:

$-3n+7r=14$

$7r=3n+14$

$r=\frac{3n+14}{7}.$

Since $r$ must be an integer, $3n+7$ and $3n+14$ must both be divisible by $7$.

Now,

$3n+7\equiv 0\, (\bmod \, 7)\Rightarrow 3n\equiv 0\, (\bmod \, 7).$

Since $3$ and $7$ are coprime,

$n\equiv 0\, (\bmod \, 7).$

So let

$n=7m.$

Then the two corresponding values of $r$ are

For ${x}^{7}$:

${r}_{1}=\frac{3(7m)+7}{7}=3m+1.$

For ${x}^{14}$:

${r}_{2}=\frac{3(7m)+14}{7}=3m+2.$

Hence the coefficients are

$(\frac{n}{3m+1})(-1{)}^{3m+1}\, \, (\frac{n}{3m+2})(-1{)}^{3m+2}.$

Their sum is zero, so

$(\frac{n}{3m+1})(-1{)}^{3m+1}+(\frac{n}{3m+2})(-1{)}^{3m+2}=0.$

Taking common sign,

$-(\frac{n}{3m+1})+(\frac{n}{3m+2})=0$

or

$(\frac{n}{3m+2})=(\frac{n}{3m+1}).$

Using the NCERT result:

$(\frac{n}{r})=(\frac{n}{r+1})\Rightarrow r+1=\frac{n}{2}$

or equivalently,

$2r+1=n.$

Here,

$r=3m+1.$

So,

$2(3m+1)+1=7m$

$6m+2+1=7m$

$6m+3=7m$

$m=3.$

Therefore,

$n=7m=21.$

Hence, the value of $n$ is

$21.$

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