How do you use the binomial formula to find the coefficient of the $ {z^{19}}{q^2}$ term in the expansion of $ {\left( {z + 2q} \right)^{21}}$ ?

Hint: We have been given a binomial expression raised to the power $ 21$ . We have to use the binomial expansion formula to expand the series. Then from the expanded series we have to find the coefficient of the term containing the expression $ {z^{19}}{q^2}$ . Binomial series expansion is given as,
$ {\left( {a + b} \right)^n} = \sum\limits_{r = 0}^n {{}^n{C_r}{a^{n - r}}{b^r}} $

Complete step by step solution:
We have been given an expression $ {\left( {z + 2q} \right)^{21}}$ . We have to use the binomial series formula and find the coefficient of the term containing the expression $ {z^{19}}{q^2}$ .
For a binomial containing terms $ a$ and $ b$ and raised to the power $ n$ , the series expansion is given as,
$ {\left( {a + b} \right)^n} = \sum\limits_{r = 0}^n {{}^n{C_r}{a^{n - r}}{b^r}} $
The series contains $ \left( {n + 1} \right)$ terms. The $ {\left( {r + 1} \right)^{th}}$ term is given as $ {}^n{C_r}{a^{n - r}}{b^r}$ .
We can see that in the $ {\left( {r + 1} \right)^{th}}$ the power of first term is $ n - r$ and the power of second term is $ r$ .
Using the binomial formula we can write the given expression as,
$ {\left( {z + 2q} \right)^{21}} = \sum\limits_{r = 0}^{21} {{}^{21}{C_r}{z^{21 - r}}{{\left( {2q} \right)}^r}} $
We have to find the coefficient of the term having power of $ z$ as $ 19$ and power of $ q$ as $ 2$ .
This means $ r = 2$ .
So we have to find the coefficient of $ \left( {r + 1} \right) = 2 + 1 = {3^{rd}}$ term.
The $ {3^{rd}}$ term is given as,
$ {}^{21}{C_2}{z^{21 - 2}}{\left( {2q} \right)^2} = {}^{21}{C_2}.{\left( 2 \right)^2}{z^{19}}{q^2}$
The coefficient of the term $ {}^{21}{C_2}.{\left( 2 \right)^2}{z^{19}}{q^2}$ is $ {}^{21}{C_2}.{\left( 2 \right)^2}$ .
$ {}^{21}{C_2}.{\left( 2 \right)^2} = 4.\dfrac{{21!}}{{\left( {21 - 2} \right)!2!}} = 4.\dfrac{{21 \times 20 \times 19!}}{{19!2!}} = 21 \times 20 \times 2 = 840$
Hence, the coefficient of the $ {z^{19}}{q^2}$ term is $ 840$ .
So, the correct answer is “840”.

Note: We used the binomial formula to find the general term of the series. The expanded series contains $ n + 1$ terms. When we have to calculate the coefficient of term with power of second variable $ r$ we have to see the $ {\left( {r + 1} \right)^{th}}$ term. We evaluate the constant term of the term to find the coefficient.