What is the coefficient a_6 in (1+x)^21+cdots+(1+x)^30 ?

Aug 13, 2017

${\text{_21C_6+""_22C_6+...+}}_{30} {C}_{6.}$

Explanation:

Recall that, (1+x)^n=sum_(r=0)^(r=n)""_nC_rx^r.

Therefore, the co-eff. of ${x}^{r} \text{ in the expansion of } {\left(1 + x\right)}^{n}$ is ""_nC_r.

Hence, the co-effs. of ${x}^{6}$ in ${\left(1 + x\right)}^{n} , w h e r e , n = 21 , 22 , \ldots , 30 ,$

are, ${\text{_21C_6, ""_22C_6, ..., }}_{30} {C}_{6} ,$ resp.

$\therefore$The Reqd. Co-eff.=${\text{_21C_6+""_22C_6+...+}}_{30} {C}_{6.}$

Aug 13, 2017

See below.

Explanation:

${\left(1 + x\right)}^{21} + \cdots + {\left(1 + x\right)}^{30} = {\left(1 + x\right)}^{21} \left(1 + \left(1 + x\right) + \cdots + {\left(1 + x\right)}^{9}\right) =$

$= {\left(1 + x\right)}^{21} \frac{{\left(1 + x\right)}^{10} - 1}{\left(1 + x\right) - 1} = \frac{1}{x} \left({\left(1 + x\right)}^{31} - {\left(1 + x\right)}^{21}\right)$

Now

${\left(1 + x\right)}^{n} = {\sum}_{k = 1}^{n} \left(\begin{matrix}k \\ n\end{matrix}\right) {x}^{k}$ then

a_6 = ((7),(31))-((7),(21)) = (31!)/(7! (31 - 7)!) - (21!)/(7! (21 - 7)!) = 2513295

Aug 13, 2017

$C o e f \left({x}^{6}\right) = 2513295$

Explanation:

We seek the coefficient of ${x}^{6}$ in:

$S = {\left(1 + x\right)}^{21} + {\left(1 + x\right)}^{22} + \ldots + {\left(1 + x\right)}^{30}$

Consider the sum

${S}_{n} = {\left(1 + x\right)}^{0} + {\left(1 + x\right)}^{1} + {\left(1 + x\right)}^{2} + \ldots + {\left(1 + x\right)}^{n}$

These term for a GP with $\left(n + 1\right)$ terms:

$a = 1$
$r = 1 + x$

And so we can use the GP sum formula to evaluate the sum of the first $n$ terms:

$S = a \frac{1 - {r}^{n}}{1 - r}$
$\setminus \setminus = \frac{1 - {\left(1 + x\right)}^{n}}{1 - \left(1 + x\right)}$
$\setminus \setminus = \frac{{\left(1 + x\right)}^{n} - 1}{x}$

Then we can find $S$ using:

$S = {S}_{31} - {S}_{21}$

$\setminus \setminus = \frac{{\left(1 + x\right)}^{31} - 1}{x} - \frac{{\left(1 + x\right)}^{21} - 1}{x}$

$\setminus \setminus = {\left(1 + x\right)}^{31} / x - {\left(1 + x\right)}^{21} / x$

We can determine the coefficient of ${x}^{r}$ in a binomial expansion ${\left(1 + x\right)}^{n}$ using the combination:

 ""_nC^r = ( (n), (r) ) = (n!)/((n-r)!r!)

We can evaluate the combination directly using factorials, or more likely using a calculators in-built function for combinations.

As dividing by $x$ will reduce the power of $x$ in the binomial expansion then the coefficient of ${x}^{6}$ that we seek is given by

$C o e f \left({x}^{6}\right) = C o e f \left({x}^{7}\right) \left\{{\left(1 + x\right)}^{31}\right\} - C o e f \left({x}^{7}\right) \left\{{\left(1 + x\right)}^{21}\right\}$
$\text{ } = \left(\begin{matrix}31 \\ 7\end{matrix}\right) - \left(\begin{matrix}21 \\ 7\end{matrix}\right)$
$\text{ } = 26295755 - 116280$
$\text{ } = 2513295$