Evaluate $\sum_{k=1}^{99}C(100,k)4^{k+2}$

Question

Evaluate the following sum and show your work. Leave expressions of the form ݉$m^n$ in your answer without attempting to evaluate them. Solutions that rely solely on calculator computation of the 100 terms of the sum are unacceptable. $$\sum_{k=1}^{99} C(100,k)4^{k+2}$$

Please verify my answer below, I would greatly appreciate any suggestions to improve the readability or quality of my work.

Answer 1

The sum to be computed is $$\sum_{k=1}^{99} {100 \choose k}4^{k+2}$$

The Binomial theorem states that

$$(x+y)^n=\sum_{k=0}^n { n \choose k} x^k y^{n-k} =x^n+y^n+ \sum_{k=1}^{n-1} {n \choose k}x^ky^{n-k}$$

For $n=100$, $x=4$ and $y=1$, this yields $$\sum_{k=1}^{99} {100 \choose k}4^{k+2}=16\sum_{k=1}^{99} {100 \choose k}4^{k}1^{n-k}=16\left((4+1)^{100}-4^{100}-1\right)$$ that is, $$\sum_{k=1}^{99} {100 \choose k}4^{k+2}=16\left(5^{100}-4^{100}-1\right)$$

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To verify the accuracy of my answer, I plugged the original summation into a calculator. $$\sum_{k=1}^{99} {100 \choose k}4^{k+2} =126217744809650880157732726036574403270254427977088432862719943541427968$$ Then compared it to my answer, $$16[5^{100}-4^{100}-1] =126217744809650880157732726036574403270254427977088432862719943541427968$$

Answer 2

$$\sum_{k=1}^{99} {100 \choose k}4^{k+2}$$

$$=4^2\left[\sum_{k=1}^{99} {100 \choose k}4^k\right]$$

$$=4^2\left[-{100 \choose 0}4^0-{100 \choose 100}4^{100}+\sum_{k=0}^{100} {100 \choose k}4^k\right]$$

Using Binomial Expansion formula, $$\sum_{k=0}^{100} {100 \choose k}4^k=(1+4)^{100}$$