Question

please solve this question ..........please......​

Answer 1

Answer:

To Prove ∑nr=03r×Crn=4n by binomial expansion

(a+b)n=C0nanb0+C1nan−1b1....Cnna0bn=

∑nr=0Crn×an−r×br

if

we compare the two terms

∑nr=03r×Crn=∑nr=0Crn×an−r×br

hence if we make a=1 and b=3

we get ∑nr=0Crn×1n−r×3r

which is same as given term,

∑nr=03r×Crn

putting the same values in (a+b)n=(1+3)n=4n