Question

Reason
$$^nC_0+^nC_4+^nC_7+.......=2^{n/2} sin (\dfrac{n \pi}{4})$$
Assertion
Consider an event for which probability of success is 1/2.
Probability that in n trials,there are r success where r-4K and k is an integer is $$\dfrac{1}{4}+\dfrac{1}{2^{n/2+1}}cos (\dfrac{n\pi}{4})$$

Answer 1

Correct option is C. Assertion is correct but Reason is incorrect
According to statement 1, the required probability is
$$^nC_0(\dfrac{1}{2})^n+^nC_4(\dfrac{1}{2})^4(\dfrac{1}{2})^{n-4}+^nC_8(\dfrac{1}{2})^8(\dfrac{1}{2})^{n-8}+.....$$
$$=(^nC_0+^nC_4+^nC_8+.....)(\dfrac{1}{2})^n$$
Now consider the binomial expansion,
$$(1+x)^n=^nC_0+^nC_1x+^nC_2x^2+....$$
Putting x +i,where $$i=\sqrt{-1}$$,we get
$$(1+i)^n=(^nC_0-^nC_2+^nC_4-....)+i(^nC_1-^nc_3+^nC_5-......)$$
$$\Rightarrow [\sqrt{2}(cos \dfrac{\pi}{4})+i\,sin\,\dfrac{\pi}{4}]^n=(^nC_0-^nC_2+^nC_4-....)+i(^nC_1-^nC_3+^nC_5-.......)$$
$$\Rightarrow ^nC_0-^nC_2+^nC_4-...=2^{n/2}cos \dfrac{n \pi}{4}$$
Hence, the required probability is
$$\dfrac{1}{4}+\dfrac{1}{2^{n/2=1}}cos(\dfrac{n \pi}{4})$$