SOLUTION: Find \sum_{k = 0}^{10} (k + 3) \cdot 2^k \cdot (k

SOLUTION: Find \sum_{k = 0}^{10} (k + 3) \cdot 2^k \cdot (k - 3)

Question 1209813: Find
\sum_{k = 0}^{10} (k + 3) \cdot 2^k \cdot (k - 3)
Answer by CPhill(1987) (Show Source): You can put this solution on YOUR website!
Let $S = \sum_{k=0}^{10} (k+3)(k-3)2^k = \sum_{k=0}^{10} (k^2 - 9)2^k$.
We can split the sum into two parts:
$S = \sum_{k=0}^{10} k^2 2^k - 9 \sum_{k=0}^{10} 2^k$.
First, let's evaluate $\sum_{k=0}^{10} 2^k$. This is a geometric series with $a=1$, $r=2$, and $n=11$ terms.
$\sum_{k=0}^{10} 2^k = \frac{1(2^{11}-1)}{2-1} = 2^{11} - 1 = 2048 - 1 = 2047$.
Now, let's evaluate $\sum_{k=0}^{10} k^2 2^k$.
Let $f(x) = \sum_{k=0}^{10} x^k = \frac{x^{11}-1}{x-1}$.
Then $xf'(x) = \sum_{k=0}^{10} kx^k$.
And $x(xf'(x))' = \sum_{k=0}^{10} k^2 x^k$.
$f'(x) = \frac{11x^{10}(x-1) - (x^{11}-1)}{(x-1)^2} = \frac{10x^{11}-11x^{10}+1}{(x-1)^2}$.
$xf'(x) = \frac{10x^{12}-11x^{11}+x}{(x-1)^2}$.
$(xf'(x))' = \frac{(120x^{11}-121x^{10}+1)(x-1)^2 - (10x^{12}-11x^{11}+x)(2(x-1))}{(x-1)^4} = \frac{(120x^{11}-121x^{10}+1)(x-1) - 2(10x^{12}-11x^{11}+x)}{(x-1)^3}$.
$x(xf'(x))' = \frac{(120x^{12}-121x^{11}+x)(x-1) - 2x(10x^{12}-11x^{11}+x)}{(x-1)^3}$.
Now, let $x=2$:
$\sum_{k=0}^{10} k 2^k = \frac{10\cdot 2^{12} - 11\cdot 2^{11} + 2}{(2-1)^2} = 10\cdot 4096 - 11\cdot 2048 + 2 = 40960 - 22528 + 2 = 18434$.
$\sum_{k=0}^{10} k^2 2^k = \frac{(120\cdot 2^{12} - 121\cdot 2^{11} + 2)(2-1) - 2(10\cdot 2^{12} - 11\cdot 2^{11} + 2)}{(2-1)^3} = 120\cdot 4096 - 121\cdot 2048 + 2 - 2(10\cdot 4096 - 11\cdot 2048 + 2) = 491520 - 248000+2-2(40960-22528+2)= 491522-248000-2(18434)= 243522-36868 = 206654$.
$S = \sum_{k=0}^{10} k^2 2^k - 9 \sum_{k=0}^{10} 2^k = 206654 - 9(2047) = 206654 - 18423 = 188231$.
Final Answer: The final answer is $\boxed{188231}$

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