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 $a={\left(1+x\right)}^{6 }and r=\left(1+x\right) and n=10$
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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