Solveeit Logo
Login

Question

Question: How do you use the binomial formula to find the coefficient of the \( {z^{19}}{q^2}\) term in the ex...

How do you use the binomial formula to find the coefficient of the z19q2 {z^{19}}{q^2} term in the expansion of (z+2q)21 {\left( {z + 2q} \right)^{21}} ?

Explanation

Solution

Hint : We have been given a binomial expression raised to the power 21 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 z19q2 {z^{19}}{q^2} . Binomial series expansion is given as,
(a+b)n=∑r=0nnCran−rbr{\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 (z+2q)21 {\left( {z + 2q} \right)^{21}} . We have to use the binomial series formula and find the coefficient of the term containing the expression z19q2 {z^{19}}{q^2} .
For a binomial containing terms a a and b b and raised to the power n n , the series expansion is given as,
(a+b)n=∑r=0nnCran−rbr{\left( {a + b} \right)^n} = \sum\limits_{r = 0}^n {{}^n{C_r}{a^{n - r}}{b^r}}
The series contains (n+1) \left( {n + 1} \right) terms. The (r+1)th {\left( {r + 1} \right)^{th}} term is given as nCran−rbr {}^n{C_r}{a^{n - r}}{b^r} .
We can see that in the (r+1)th {\left( {r + 1} \right)^{th}} the power of first term is n−r n - r and the power of second term is r r .
Using the binomial formula we can write the given expression as,
(z+2q)21=∑r=02121Crz21−r(2q)r{\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 z as 19 19 and power of q q as 2 2 .
This means r=2 r = 2 .
So we have to find the coefficient of (r+1)=2+1=3rd \left( {r + 1} \right) = 2 + 1 = {3^{rd}} term.
The 3rd {3^{rd}} term is given as,
21C2z21−2(2q)2=21C2.(2)2z19q2{}^{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 21C2.(2)2z19q2 {}^{21}{C_2}.{\left( 2 \right)^2}{z^{19}}{q^2} is 21C2.(2)2 {}^{21}{C_2}.{\left( 2 \right)^2} .
21C2.(2)2=4.21!(21−2)!2!=4.21×20×19!19!2!=21×20×2=840{}^{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 z19q2 {z^{19}}{q^2} term is 840 840 .
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 n + 1 terms. When we have to calculate the coefficient of term with power of second variable r r we have to see the (r+1)th {\left( {r + 1} \right)^{th}} term. We evaluate the constant term of the term to find the coefficient.