Question

please solve this question

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Answer 1

Answer:

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

To Prove ∑nr=03r×Crn=4n by binomial expansion(a+b)n=C0nanb0+C1nan−1b1....Cnna0bn=

To Prove ∑nr=03r×Crn=4n by binomial expansion(a+b)n=C0nanb0+C1nan−1b1....Cnna0bn=∑nr=0Crn×an−r×br

To Prove ∑nr=03r×Crn=4n by binomial expansion(a+b)n=C0nanb0+C1nan−1b1....Cnna0bn=∑nr=0Crn×an−r×brif

To Prove ∑nr=03r×Crn=4n by binomial expansion(a+b)n=C0nanb0+C1nan−1b1....Cnna0bn=∑nr=0Crn×an−r×brifwe compare the two terms

To Prove ∑nr=03r×Crn=4n by binomial expansion(a+b)n=C0nanb0+C1nan−1b1....Cnna0bn=∑nr=0Crn×an−r×brifwe compare the two terms∑nr=03r×Crn=∑nr=0Crn×an−r×br

To Prove ∑nr=03r×Crn=4n by binomial expansion(a+b)n=C0nanb0+C1nan−1b1....Cnna0bn=∑nr=0Crn×an−r×brifwe compare the two terms∑nr=03r×Crn=∑nr=0Crn×an−r×brhence if we make a=1 and b=3

To Prove ∑nr=03r×Crn=4n by binomial expansion(a+b)n=C0nanb0+C1nan−1b1....Cnna0bn=∑nr=0Crn×an−r×brifwe compare the two terms∑nr=03r×Crn=∑nr=0Crn×an−r×brhence if we make a=1 and b=3we get ∑nr=0Crn×1n−r×3r

To Prove ∑nr=03r×Crn=4n by binomial expansion(a+b)n=C0nanb0+C1nan−1b1....Cnna0bn=∑nr=0Crn×an−r×brifwe compare the two terms∑nr=03r×Crn=∑nr=0Crn×an−r×brhence if we make a=1 and b=3we get ∑nr=0Crn×1n−r×3rwhich is same as given term,

To Prove ∑nr=03r×Crn=4n by binomial expansion(a+b)n=C0nanb0+C1nan−1b1....Cnna0bn=∑nr=0Crn×an−r×brifwe compare the two terms∑nr=03r×Crn=∑nr=0Crn×an−r×brhence if we make a=1 and b=3we get ∑nr=0Crn×1n−r×3rwhich is same as given term,∑nr=03r×Crn

To Prove ∑nr=03r×Crn=4n by binomial expansion(a+b)n=C0nanb0+C1nan−1b1....Cnna0bn=∑nr=0Crn×an−r×brifwe compare the two terms∑nr=03r×Crn=∑nr=0Crn×an−r×brhence if we make a=1 and b=3we get ∑nr=0Crn×1n−r×3rwhich is same as given term,∑nr=03r×Crnputting the same values in (a+b)n=(1+3)n=4n