If (1-x^3)^10= _r=0^10a_rx^r(1-x)^30-2r, then 9a_9a_10 is equal to ____ .

Question

If ${(1-{x}^{3})}^{10}=\sum _{r=0}^{10}{a}_{r}{x}^{r}(1-x{)}^{30-2r}$, then $\frac{9{a}_{9}}{{a}_{10}}$ is equal to $____$ .

Answer: 30

Accepted Answer

Given

$(1-{x}^{3}{)}^{10}=\sum _{10}^{r=0}{a}_{r}{x}^{r}(1-x{)}^{30-2r}.$

We have to find

$\frac{9{a}_{9}}{{a}_{10}}.$

Let us simplify the left side first.

Since

$1-{x}^{3}=(1-x)(1+x+{x}^{2}),$

so

$(1-{x}^{3}{)}^{10}=(1-x{)}^{10}(1+x+{x}^{2}{)}^{10}.$

Also,

$1+x+{x}^{2}=\frac{1-{x}^{3}}{1-x}.$

Now notice that

$1+x+{x}^{2}=(1-x{)}^{-2}(x+(1-x{)}^{2})?$

This is not the most useful form here. So let us proceed in a better NCERT-style way.

We are given

$(1-{x}^{3}{)}^{10}=\sum _{10}^{r=0}{a}_{r}{x}^{r}(1-x{)}^{30-2r}.$

Take out $(1-x{)}^{10}$ from the left side:

$(1-{x}^{3}{)}^{10}=(1-x{)}^{10}(1+x+{x}^{2}{)}^{10}.$

Hence

$(1+x+{x}^{2}{)}^{10}=\sum _{10}^{r=0}{a}_{r}{x}^{r}(1-x{)}^{20-2r}.$

Now observe that

$1+x+{x}^{2}=(1-x{)}^{2}+3x.$

Because

$(1-x{)}^{2}+3x=1-2x+{x}^{2}+3x=1+x+{x}^{2}.$

Therefore

$(1+x+{x}^{2}{)}^{10}=((1-x{)}^{2}+3x{)}^{10}.$

Using binomial theorem,

$((1-x{)}^{2}+3x{)}^{10}=\sum _{10}^{r=0}(\frac{10}{r})(3x{)}^{r}((1-x{)}^{2}{)}^{10-r}.$

So

$(1+x+{x}^{2}{)}^{10}=\sum _{10}^{r=0}(\frac{10}{r}){3}^{r}{x}^{r}(1-x{)}^{20-2r}.$

Comparing this with

$(1+x+{x}^{2}{)}^{10}=\sum _{10}^{r=0}{a}_{r}{x}^{r}(1-x{)}^{20-2r},$

we get

${a}_{r}=(\frac{10}{r}){3}^{r}.$

Hence

${a}_{9}=(\frac{10}{9}){3}^{9}=10\cdot {3}^{9},$

and

${a}_{10}=(\frac{10}{10}){3}^{10}={3}^{10}.$

Therefore

$\frac{9{a}_{9}}{{a}_{10}}=9\cdot \frac{10\cdot {3}^{9}}{{3}^{10}}=9\cdot \frac{10}{3}=30.$

So the required value is

$30.$

Solutions · 1