Solution: 2013 Fall Midterm - 9

Author: Pat Morin

Question

Which of the following is true?

(a)

$\sum_{k=0}^{n} 4^{k}5^{n-k}{{n}\choose{k}} = 20^{n}$

(b)

$\sum_{k=0}^{n} 5^{k}{{n}\choose{k}} = 6^{n}$

(c)

$\sum_{k=0}^{n} 5^{k}{{n}\choose{k}} = 5^{n}$

(d)

$\sum_{k=0}^{n} 4^{n-k}5^{k}{{n}\choose{k}} = 8^{n}$

It’s obvious which one is right BUT let’s make it clear by going through every single one of them just in case and seeing what they come out to

$\sum_{k=0}^{n} 4^{k}5^{n-k}{{n}\choose{k}} = 20^{n}$

${(4+5)}^n = {20}^n$

$9^n = {20}^n$

$\sum_{k=0}^{n} 5^{k}{{n}\choose{k}} = 6^{n}$

$\sum_{k=0}^{n} \binom{n}{k} (1)^{n-k} (5^k) = 6^n$

${(1+5)}^n = 6^n$

$ 6^n = 6^n$

$\sum*{k=0}^{n} 5^{k}{{n}\choose{k}} = 5^{n}$

$\sum*{k=0}^{n} \binom{n}{k} (1)^{n-k} (5^k) = 5^n$

${(1+5)}^n = 5^n$

$ 6^n = 5^n$

$\sum_{k=0}^{n} 4^{n-k}5^{k}{{n}\choose{k}} = 8^{n}$

$ {(4+5)}^n= 8^{n}$

$ 9^n= 8^{n}$