Proving the Binomial Theorem: Simplifying 3^k C(n,k) = 2^2n

[jenny Downer]

Use the Binomial Theorem to show that n summation k-0 3^k C(n,k) = 2^2n

hint of the question is 3^k C(n,k) = 3^k 1^n-k C(n,k)

 

[Dick]

Think of applying the binomial theorem to the expansion of (1+3)^n.

 

[Architeuthis Dux]

= 3^k 1^n-k = 2^2n

The Binomial Theorem is a fundamental concept in mathematics that allows us to expand the power of a binomial expression. In this case, we are interested in proving the equation 3^k C(n,k) = 2^2n, which can be rewritten as 3^k 1^(n-k) C(n,k) = 2^2n. This can be further simplified as 3^k C(n,k) = 2^2n, since 1^(n-k) is equal to 1.

To understand how this equation is derived, we must first understand the Binomial Theorem. It states that for any positive integers a and b, and any non-negative integer n, the expansion of (a+b)^n can be written as the sum of the coefficients of each term multiplied by a and b raised to various powers. This can be represented as (a+b)^n = Σ(n, k=0) C(n,k) a^(n-k) b^k, where C(n,k) represents the binomial coefficient.

In our case, we have (3+1)^n = Σ(n, k=0) C(n,k) 3^(n-k) 1^k = Σ(n, k=0) C(n,k) 3^(n-k). However, we are only interested in the terms where k=0 and k=n, as all other terms will have a coefficient of 0 when multiplied by 3^(n-k). So, we can rewrite the equation as (3+1)^n = C(n,0) 3^n + C(n,n) 3^0 = 3^n + 1.

Next, we can use the fact that C(n,0) = C(n,n) = 1, to further simplify the equation as (3+1)^n = 2^n. By substituting 3^k C(n,k) = 2^2n, we get (3+1)^n = 3^k C(n,k) = 2^2n = 2^n. This proves that 3^k C(n,k) = 2^2n is a valid equation.

Moreover, we can show that the summation of all the terms in the expansion of (3+1)^