Let $$1+x+x2100=$$∑$$r=0200$$ arxr and $$a=$$∑$$r=0200$$ ar then value of ∑$$r=1200$$ $$rar25a$$ $$is-$$

$$a=$$∑$$r=0200$$ $$ar=$$∑$$r=0200$$ $$ar(1)r=3100Differentiating$$ (1), we $$get1001+x+x299(1+2x)=$$∑$$r=1200$$ rarxr−$$1put$$ x $$=$$ $$1$$, to $$obtain100(3)99(3)=$$∑$$r=1200$$ rar⇒ ∑$$r=1200$$ $$rar25a=100a25a=4$$

Binomial theorem for positive integral Index

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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-

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

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Binomial theorem for positive integral Index

Remember concepts with our Masterclasses.

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

a=∑r=0200 ar=∑r=0200 ar(1)r=3100Differentiating (1), we get1001+x+x299(1+2x)=∑r=1200 rarxr−1put x = 1, to obtain100(3)99(3)=∑r=1200 rar⇒ ∑r=1200 rar25a=100a25a=4

Ready to Test Your Skills?

free study material

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-