Question

What is the coefficient of ${x^{2017}}$ in $\sum\limits_{r = 0}^{2020} {{}^{2020}{C_r}} {\left( {x - 2018} \right)^{2020 - r}}{\left( {2017} \right)^r}$ ?

Answer 1

Hint: Start with expanding the given summation $\sum\limits_{r = 0}^{2020} {{}^{2020}{C_r}} {\left( {x - 2018} \right)^{2020 - r}}{\left( {2017} \right)^r}$ . Now write the binomial theorem, i.e. expansion of ${\left( {a + b} \right)^n}$ . Now compare both the equations and find the value of given summation in a ${\left( {a + b} \right)^n}$ form. Now again use the binomial theorem to expand the new expression. Find the term containing ${x^{2017}}$ in the expansion and write its coefficient.

Complete step-by-step answer:
Here in this problem, we are given with an expression $\sum\limits_{r = 0}^{2020} {{}^{2020}{C_r}} {\left( {x - 2018} \right)^{2020 - r}}{\left( {2017} \right)^r}$ and we need to find the coefficient of ${x^{2017}}$ in the expansion of the given summation.
The summation given in the expression can be expanded as:
$\sum\limits_{r = 0}^{2020} {{}^{2020}{C_r}} {\left( {x - 2018} \right)^{2020 - r}}{\left( {2017} \right)^r} = {}^{2020}{C_0}{\left( {x - 2018} \right)^{2020 - 0}}{\left( {2017} \right)^0} + {}^{2020}{C_1}{\left( {x - 2018} \right)^{2020 - 1}}{\left( {2017} \right)^1} + $$...... + {}^{2020}{C_{2019}}{\left( {x - 2018} \right)^{2020 - 2019}}{\left( {2017} \right)^{2019}} + {}^{2020}{C_{2020}}{\left( {x - 2018} \right)^{2020 - 2020}}{\left( {2017} \right)^{2020}}$
Now let’s further solve the above expansion
$ \Rightarrow \sum\limits_{r = 0}^{2020} {{}^{2020}{C_r}} {\left( {x - 2018} \right)^{2020 - r}}{\left( {2017} \right)^r} = {}^{2020}{C_0}{\left( {x - 2018} \right)^{2020}}{\left( {2017} \right)^0} + {}^{2020}{C_1}{\left( {x - 2018} \right)^{2019}}\left( {2017} \right) + $$...... + {}^{2020}{C_{2019}}{\left( {x - 2018} \right)^1}{\left( {2017} \right)^{2019}} + {}^{2020}{C_{2020}}{\left( {x - 2018} \right)^0}{\left( {2017} \right)^{2020}}$
We can now use the identity ${a^0} = 1{\text{ and }}{a^1} = a$ in the above series as:
$ \Rightarrow \sum\limits_{r = 0}^{2020} {{}^{2020}{C_r}} {\left( {x - 2018} \right)^{2020 - r}}{\left( {2017} \right)^r} = {}^{2020}{C_0}{\left( {x - 2018} \right)^{2020}} + {}^{2020}{C_1}{\left( {x - 2018} \right)^{2019}}\left( {2017} \right) + $
$...... + {}^{2020}{C_{2019}}\left( {x - 2018} \right){\left( {2017} \right)^{2019}} + {}^{2020}{C_{2020}}{\left( {2017} \right)^{2020}}$ ……….(i)
Also, we have the binomial expansion as:
\[{\left( {a + b} \right)^n} = {}^n{C_0}{a^n} + {}^n{C_1}{a^{n - 1}}b + {}^n{C_2}{a^{n - 2}}{b^2} + ...... + {}^n{C_{n - 1}}a{b^{n - 1}} + {}^n{C_n}{b^n}\] ...........(ii)
Now by comparing RHS of equation (i) and (ii), we can say that
$ \Rightarrow a = \left( {x - 2018} \right){\text{ ; }}b = 2017{\text{ ; }}n = 2020$ ......(iii)
Thus, we can also compare the LHS of the equation (i) and (ii) as
$ \Rightarrow \sum\limits_{r = 0}^{2020} {{}^{2020}{C_r}} {\left( {x - 2018} \right)^{2020 - r}}{\left( {2017} \right)^r} = {\left( {a + b} \right)^n}$
Now we can substitute the values obtained in (iii) in the above equation:
$ \Rightarrow \sum\limits_{r = 0}^{2020} {{}^{2020}{C_r}} {\left( {x - 2018} \right)^{2020 - r}}{\left( {2017} \right)^r} = {\left( {a + b} \right)^n} = {\left( {x - 2018 + 2017} \right)^{2020}}$
Therefore, for the given expression we get a simplified form as:
$ \Rightarrow \sum\limits_{r = 0}^{2020} {{}^{2020}{C_r}} {\left( {x - 2018} \right)^{2020 - r}}{\left( {2017} \right)^r} = {\left( {x - 1} \right)^{2020}}$
We just need to find the coefficient of ${x^{2017}}$ in the expansion of ${\left( {x - 1} \right)^{2020}}$. And the expansion for ${\left( {x - 1} \right)^{2020}}$ can easily be given by the use of binomial expansion (ii).
$ \Rightarrow $ For ${\left( {x - 1} \right)^{2020}}$, in identity (ii), we have: $a = x,b = - 1{\text{ and }}n = 2020$
So we can write the binomial expansion for ${\left( {x - 1} \right)^{2020}}$ using (ii), as:
\[ \Rightarrow {\left( {x - 1} \right)^{2020}} = {}^{2020}{C_0}{x^{2020}} + {}^{2020}{C_1}{x^{2020 - 1}}\left( { - 1} \right) + {}^{2020}{C_2}{x^{2020 - 2}}{\left( { - 1} \right)^2} + {}^{2020}{C_3}{x^{2020 - 3}}{\left( { - 1} \right)^3}......\]
                                                 \[...... + {}^{2020}{C_{2020 - 2}}{x^2}{\left( { - 1} \right)^{2020 - 2}} + {}^{2020}{C_{2020 - 1}}x{\left( { - 1} \right)^{2020 - 1}} + {}^{2020}{C_{2020}}{\left( { - 1} \right)^{2020}}\]
Thus, from the above expansion, we can see that each term contains $'x'$ raised to a different power from $0 - 2020$ .
Only the fourth term from the beginning which contains ${x^{2017}}$ in it and thus the coefficient of this power can be evaluated by evaluating that term
$ \Rightarrow $ The fourth term of the expansion $ = {}^{2020}{C_3}{x^{2020 - 3}}{\left( { - 1} \right)^3} = {}^{2020}{C_3}{x^{2017}}{\left( { - 1} \right)^3}$
Therefore, the coefficient of ${x^{2017}}$ $ = {}^{2020}{C_3}{\left( { - 1} \right)^3} = - {}^{2020}{C_3}$

Note: Be careful with the signs and parenthesis while expanding the binomial expression. In mathematics, a coefficient is a multiplicative factor in some term of a polynomial, a series, or any expression; it is usually a number, but maybe any expression. Also remember the property of combinations that says: ${}^n{C_r} = {}^n{C_{n - r}}$ . The answer that we got, i.e. $ - {}^{2020}{C_3}$, can also be written as $ - {}^{2020}{C_{2020 - 3}} = - {}^{2020}{C_{2017}}$ and it will still be the same numerically.