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 the given expansion, we get100(1+x+x)299(1+2x)=∑r=1200 rarxr−1Put x=1, to obtain100(3)99(3)=∑r=1200 rar⇒∑r=1200 rar25a=100a25a=4

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 the given expansion, we get100(1+x+x)299(1+2x)=∑r=1200 rarxr−1Put x=1, to obtain100(3)99(3)=∑r=1200 rar⇒∑r=1200 rar25a=100a25a=4

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

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

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

answer is B.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

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

Differentiating the given expansion, 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

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