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$\sum_{r=0}^{300} a_{r} x^{r}=\left(1+x+x^{2}+x^{3}\right)^{100}$ . If $a=\sum_{r=0}^{300} a_{r}$ then $\sum_{r=0}^{300} r \cdot a_{r}$ is equal to
Option: 1
300 a

Option: 2
100 a

Option: 3
150 a

Option: 4
75 a

Answers (1)

best_answer

(C) $\sum_{r=0}^{300} a_{r} r^{r}=\left(1+x+x^{2}+x^{3}\right)^{100}$

Clearly, ' $a_{r}$ ' is the coefficient of $x^{r}$ in the expansion of $\left(1+x+x^{2}+x^{3}\right)^{100}$ .

Replacing $x \rightarrow \frac{1}{x}$ in the given equation, we get

$\sum_{r=0}^{300} a_{r}\left(\frac{1}{x}\right)^{r}=\frac{1}{x^{300}}\left(x^{3}+x^{2}+x+1\right)^{100} \\$

$\Rightarrow \sum_{r=0}^{300} a_{r} x^{300-r}=\left(1+x+x^{2}+x^{3}\right)^{100}$

Here $a_{r}$ represents of coefficient. of $x^{300-r}$ in $\left(1+x+x^{2}+x^{3}\right)^{100}$

Thus, $a_{r}=a_{300-r}$

Let, $\quad I=\sum_{r=0}^{300} r \cdot a_{r}=\sum_{r=0}^{300}(300-r) a_{300-r}$

$\sum_{r=0}^{300}(300-r) a_{r} \quad \sum_{r=0}^{300} a_{r}-\sum_{r=0}^{300} r \cdot a_{r} \\$

$\Rightarrow 2l=300 a \\$

$\Rightarrow I=150 a$

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