Question

The coefficient of ${x^{53}}$ in the expansion of $\sum\limits_{M = 0}^{100} {} $$^{100}Cm{\left( {x - 3} \right)^{100 - m}},{2^m}$ is
(a) $^{100}{C_{47}}$
(b) $^{100}{C_{53}}$
(c) $^{ - 100}{C_{53}}$
(d) None of these

Answer 1

Hint: In this question use the method of expansion coefficient and then solve it with the series. This problem can be solved by the binomial expansion..as the above is a sum of series it can be written in the form of binomial expansion. so we can solve the problem.


Complete step by step solution:
Now we solve the problem by binomial expansion.
We know that
${\left( {x + y} \right)^n}{ = ^n}{C_0}{x^n}{ + ^x}{C_1}{x^{n - 1}}y{ + ^n}{C_2}{x^{n - 2}}{y^2} + ...{ + ^x}{C_{n - 1}}x{y^{n - 1}}{ + ^x}{C_n}{y^x}$
${\left( {x + y} \right)^n} = \sum\limits_{r = 0}^n {^n{C_r}{x^{n - r}}{y^r}} $
According to the equation $\sum\limits_{1m = 0}^{100} {^{100}Cm} {\left( {x - 3} \right)^{10 - m}}{2^m}$
This can be written as—
     ${\left[ {\left( {x - 3} \right) + 2} \right]^{100}}$
     ${\left( {x - 1} \right)^{100}} \Rightarrow {\left( {1 - x} \right)^{100}}$
We know,
     ${\left( {1 - x} \right)^n}{ = ^n}{C_6}{ - ^n}{C_1}x{ + ^x}{C_2}{x^2}.... + {\left( { - 1} \right)^{rn}}{C_r}{x^r}$
     ${\left( {1 - x} \right)^{100}}{ = ^{100}}{C_0}{x^{100}} - 100{C_1}{x^{99}}{ + ^{100}}{C_2}{x^{98}}....{ + ^{100}}{C_{100}}$
${x^{53}}$ will occur in ${T_{54}}$
${T_{54}} = {T_{53 + 1}}{ = ^{100}}{C_{53}}{\left( { - x} \right)^{53}}$
     ${ = ^{ - 100}}{C_{53}}{x^{53}}$
Coefficient of ${x^{53}}$ is $^{ - 100}{C_{53}}$
Correct answer is \[\left( d \right)\]


Note: Make sure to write the appropriate coefficients associated with the power.And also when applying the formula ,make sure to substitute the right values