Using the Binomial Theorem to Show 3^k C(n,k) = 3^k 1^n-k C(n,k)

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The discussion focuses on using the Binomial Theorem to demonstrate that 3^k C(n,k) equals 3^k 1^(n-k) C(n,k). It emphasizes that C(n,k) represents the binomial coefficients, which are derived from the expansion of (x + y)^n. The key insight is that when substituting x and y with specific values, the coefficients can be evaluated. The equation n sum 3^k C(n,k) = 2^(2n) is also highlighted as a significant result of this approach. Understanding these relationships is crucial for applying the Binomial Theorem effectively.
jenny Downer
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how do i use binomial to show that 3^k C(n,k) = 3^k 1^n-k C(n,k)
 
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Did you type that correctly? Right now you have stuff = stuff * 1, which is clearly correct...
 
i wanted to explain a bit more
 
It should be
n
sum 3^k C(n,k) = 2^2n
k=0

the hint is 3^k C(n,k) = 3^k 1^n-k C(n,k)
 
The reason C(n, k) are called "binomial coefficients" is that C(n, k) is the coefficient of xk in (x+ y)n What are the coefficients of xk in (3x+ y)n? What do you get if x= y= 1?
 

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