Question

I want full explanation​

Answer 1

Answer:

See below

Step-by-step explanation:

we would like to prove the following question:

$\displaystyle \sum _{r = 0} ^{n} {3}^{r} \ \binom{n}{r} = {4}^{n}$

In order to do so, recall binomial theorem:

$\displaystyle \sum _{r= 0} ^{n} {a}^{n - r} {b}^{r} \ \binom{n}{r} = (a + b {)}^{n}$

where:

• n refers to the degree of a binomial
• C(n,k) is the coefficient of the each term

Considering the given problem,if we assume a and b are 1 and 3 respectively then it can be proved easily so let's try it!

$\displaystyle \sum _{r= 0} ^{n} {1}^{n - r} {3}^{r} \ \binom{n}{r} \stackrel {?}{ = } (1 + 3 {)}^{n}$

remember that,any power to 1 always yields 1 therefore we can drop $1^{n-r}$ so, we acquire:

$\displaystyle \sum _{r= 0} ^{n} {3}^{r} \ \binom{n}{r} \stackrel {?}{ = } (1 + 3 {)}^{n}$

simplify right hand side:

$\displaystyle \sum _{r= 0} ^{n} {3}^{r} \ \binom{n}{r} \stackrel { \checkmark}{ = } (4{)}^{n}$

hence we are done!

note:${}^n C_r$ is also written as $\binom{n}{r}$