Question

$\sum _{r=0}^{300}{a}_r{x}^r=(1+x+x^2+x^3{)}^{100}$. If $a=\sum _{r=0}^{300}{a}_r$, then $\sum _{r=0}^{300}r{a}_r$ is equal to

• A. $300a$
• B. $100a$
• C. $150a$
• D. $75a$

Answer 1

The correct option is C $150a$

$\sum _{r=0}^{300}{a}_r{x}^r=(1+x+x^2+x^3{)}^{100}$
Clearly, '$a_r$' is the coefficient of $x_r$ in the expansion of $(1+x+x^2+x^3{)}^{100}$
Replacing $x$ by $\frac{1}{x}$ in the given equation, we get
$\sum _{r=0}^{300}{a}_r{\left(\frac{1}{x}\right)}^r=\frac{1}{x^{300}}(x^3+x^2+x+1{)}^{100}$
or, $\sum _{r=0}^{300}{a}_r{x}^{300-r}=(1+x+x^2+x^3{)}^{100}$

Here, $a_r$ represents the coefficients of $x^{300-r}$ in the expansion of $(1+x+x^2+x^3{)}^{100}$.
Thus, $a_r=a_{300-r}$
Let $I=\sum _{r=0}^{300}r\times a_r$
$=\sum _{r=0}^{300}(300-r)a_{300-r}$
$=\sum _{r=0}^{300}(300-r)a_r$
$=300\sum _{r=0}^{300}{a}_r-\sum _{r=0}^{300}r{a}_r$
$\Rightarrow 2I=300a$
or, $I=150a$