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I want full explanation​-example-1
User Shirkan
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10 votes

Answer:

See below

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}

User Mark Foreman
by
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