# Above is question and its solution. I didn't get the first step itself. Explain in detail

Dear student,
given series is ${\left(1+x\right)}^{6}+{\left(1+x\right)}^{7}+.............+{\left(1+x\right)}^{15}$ ..............(1)
as we can see it is an G.P
and we know that $a+ar+a{r}^{2}+a{r}^{3}+.......+a{r}^{n-1}=\frac{a\left({r}^{n}-1\right)}{r-1}$
here
now (1) will become $\frac{{\left(1+x\right)}^{6}\left\{{\left(1+x\right)}^{10}-1\right\}}{\left(1+x\right)-1}$
or $\frac{{\left(1+x\right)}^{16}-\left\{{\left(1+x\right)}^{6}\right\}}{x}$
$\frac{{\left(1+x\right)}^{16}}{x}-\frac{{\left(1+x\right)}^{6}}{x}$
to find coefficient of ${x}^{6}$,we need to find the coefficient of ${x}^{7}$ in ${\left(1+x\right)}^{16}$
we also know that coefficient of ${x}^{r}$ in ${\left(1+x\right)}^{n}$ is ${}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}}$
so coefficient of ​${x}^{7}$ is ${}^{16}{\mathrm{C}}_{7}$=${}^{16}{\mathrm{C}}_{16-7}$=${}^{16}{\mathrm{C}}_{9}$     (since ${}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}}$=${}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{n}-\mathrm{r}}$)

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They have considered the expression to be a GP. The 'variable' of GP is took as (1+x) instead of x (in normal cases).
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Yup ! Reeshab is absolutely correct ...
But rishabh its not 'variable' , its 'common ratio'
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What are you looking for?