Let 1+x+x2100=∑r=0200 arxr and a=∑r=0200 ar then value of ∑r=1200 rar25a is- Infinity Learn by Sri Chaitanya

Solution:

$a = \sum_{r = 0}^{200} a_{r} = \sum_{r = 0}^{200} a_{r} (1)^{r} = 3^{100}$

Differentiating $(1),$ we get

$100 {(1 + x + x^{2})}^{99} (1 + 2 x) = \sum_{r = 1}^{200} r a_{r} x^{r - 1}$

put $x = 1$ , to obtain

$\begin{array}{l} 100 (3)^{99} (3) = \sum_{r = 1}^{200} r a_{r} \\ \Rightarrow \sum_{r = 1}^{200} \frac{r a_{r}}{25 a} = \frac{100 a}{25 a} = 4 \end{array}$

Let ${(1 + x + x^{2})}^{100} = \sum_{r = 0}^{200} a_{r} x^{r}$ and $a = \sum_{r = 0}^{200} a_{r}$ then value of $\sum_{r = 1}^{200} \frac{r a_{r}}{25 a}$ is-

Solution:

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