Question

if r is remainder obtained on dividing (98)^5 by 12 .then coefficent of x^3 on the expansion (1+x/2)^2r

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Answer 1

Answer with explanation:

1.It is given that, r is remainder obtained on dividing $(98)^5$ by 12.

96 is divisible by 5.

So, writing, 98=96+2

With the help of binomial expansion,expanding

$(96+2)^5=_{0}^{5}\textrm{C}\times(96)^5+_{1}^{5}\textrm{C}\times(96)^4\times 2+_{2}^{5}\textrm{C}\times(96)^3\times 2^2+_{3}^{5}\textrm{C}\times(96)^2\times 2^3+_{4}^{5}\textrm{C}\times(96)^1\times 2^4+_{5}^{5}\textrm{C}\times(96)^0\times 2^5\\\\(96+2)^5=_{0}^{5}\textrm{C}\times(96)^5+_{1}^{5}\textrm{C}\times(96)^4\times 2+_{2}^{5}\textrm{C}\times(96)^3\times 2^2+_{3}^{5}\textrm{C}\times(96)^2\times 2^3+_{4}^{5}\textrm{C}\times(96)^1\times 2^4+32$

As,there are six terms in the expansion.First five terms are divisible by 12.

So, Remainder when $(98)^5$ is divided by 12=32

r=32

2.

We have to find coefficient of x³, in the expansion of $(1+\frac{x}{2})^{2r}$

Let,$t_{r+1}$ be the coefficient of x³ in the expansion of $(1+\frac{x}{2})^{2r}$.

$t_{r+1} {\text{in} (1+\frac{x}{2})^{2r}=_{r}^{2r}\textrm{C}(1)^r(\frac{x}{2})^{2r-r} \\\\\\\\ t_{r+1} =_{r}^{2r}\textrm{C}(1)^r(\frac{x^r}{2^r})$

When, r=3

that is ,

$t_{4}=_{4}^{8}\textrm{C}(1)^4(\frac{x^3}{2^3})\\\\t_{4}=_{4}^{8}\textrm{C}(1)^4(\frac{1}{2})^3=\frac{8!}{4!\times 4!}\times \frac{1}{8}=\frac{70}{8}=\frac{35}{4}$

Coefficient of $x^3=\frac{35}{4}$

Answer 2

Answer:

Step-by-step explanation:

Use binomial theorem

98^5=(96+2)^5

Now expand this using binomial theorem.

At the last when u divide the expanded eqn by 12 we get

32 ........but 32 can by either divided by 12 as 32=12×4+8

So now our remainder is 8.....

Therefore 2r= 16.

Now use binomial theorem again

16C3 (x/2)^3

= 70

This is jee main 2019 question.......