Determine the coefficient of x^{11} in the expansion of (x^2 + \frac{1}{x})^{10}.

Question:

Determine the coefficient of {eq}\,x^{11}\, {/eq} in the expansion of {eq}\;\left(x^2 + \frac{1}{x}\right)^{10} {/eq}.

Binomial Theorem:

The expansion fo the higher power numbers or the functions of the form {eq}\left(a+b\right)^n {/eq} can be done with the help of the binomial theorem and the formula of expansion by this theorem as: {eq}\left(a+b\right)^n=\sum _{i=0}^n\binom{n}{i}a^{\left(n-i\right)}b^i {/eq}

In the problem, we have to determine the coefficient of {eq}\,x^{11}\, {/eq} in the expansion of {eq}\;\left(x^2 + \frac{1}{x}\right)^{10} {/eq}.

So we use the binomial theorem:

{eq}\left(a+b\right)^n=\sum _{i=0}^n\binom{n}{i}a^{\left(n-i\right)}b^i\\ {/eq}

So the given function will be written as

{eq}\left(x^2+\frac{1}{x}\right)^{10}=\sum _{i=0}^{10}\binom{10}{i}\left(x^2\right)^{\left(10-i\right)}\left(\frac{1}{x}\right)^i~~~~~~~~~~~~~\left [ \because a=x^2,\:\:b=\frac{1}{x} \right ]\\ {/eq}

Now to get the coefficient of

{eq}x^{11} {/eq}

we have

{eq}\frac{10!}{3!\left(10-3\right)!}\left(x^2\right)^7\left(\frac{1}{x}\right)^3\\ =120x^{11} {/eq}

So this gives the coefficient as:

120