0

I'm taking a Course in Computer Science where we're having a refresher on Calculus 2 material. There is a problem that I don't understand or know how to do.

Compute the following using the shifting method: $\sum_{i=1}^h i\,2^i$

Could anyone explain to me how I would solve this problem? Sorry if I've formatted the text wrong, just let me know and i'll change it!

Martin Argerami
  • 205,756
user1781500
  • 99
  • In general, you might also find this interesting to look at: http://math.stackexchange.com/questions/30732/how-can-i-evaluate-sum-n-0-infty-n1xn – mathematics2x2life Jan 27 '14 at 19:35

2 Answers2

1

Well, let's say that $$S=\sum_{i=1}^h i2^i.$$ Now, we will let $i=j+1,$ so that $j=i-1$ and $j$ ranges from $0$ to $h-1,$ meaning that $$S=\sum_{j=0}^{h-1}(j+1)2^{j+1}\\S=\sum_{j=0}^{h-1}j2^{j+1}+\sum_{j=0}^{h-1}2^{j+1}.$$ Since $2^{j+1}=2\cdot 2^j,$ then we have $$S=2\sum_{j=0}^{h-1}j2^j+2\sum_{j=0}^{h-1}2^j,$$ and since $0\cdot 2^0=0,$ then $$S=2\sum_{j=1}^{h-1}j2^j+2\sum_{j=0}^{h-1}2^j.$$ Notice how similar that first sum looks to our first form of $S$? The only difference is that we're multiplying it by $2,$ calling our index variable by a different name, and missing the term $h2^h.$ So, to simplify our life by first making it slightly more complicated, we have $$S=2\left(-h2^h+h2^h+\sum_{j=1}^{h-1}j2^j\right)+2\sum_{j=0}^{h-1}2^j\\S=2\left(-h2^h+\sum_{j=1}^hj2^j\right)+2\sum_{j=0}^{h-1}2^j\\S=2\left(-h2^h+\sum_{i=1}^hi2^i\right)+2\sum_{j=0}^{h-1}2^j\\S=2\left(-h2^h+S\right)+2\sum_{j=0}^{h-1}2^j\\S=-2h2^h+2S+2\sum_{j=0}^{h-1}2^j\\S=-h2^{h+1}+2S+2\sum_{j=0}^{h-1}2^j.$$ Can you solve for $S$ from here?

Cameron Buie
  • 102,994
  • Where does the -h2^2 + h2^2 come from in s = 2(-h2^2 + h2^2 + summation.... – user1781500 Jan 27 '14 at 21:29
  • That is just a fancy way of adding $0$ (which is to say, changing nothing at all). Since we were missing the $h2^h$ term, one way I could get it in there without actually changing anything was to add and subtract $h2^h$ in there. Alternately, I could have added $2h2^h$ to both sides to get $$S+2h2^h=2S+2\sum_{j=0}^{h-1}2^j,$$ which gets us to the same place – Cameron Buie Jan 27 '14 at 21:34
0

Let $S=\sum_{k=1}^{h}k2^k$. If you multiply by $2$ and subtract, you get $$ S=2S-S=\sum_{k=1}^hk2^{k+1}-\sum_{k=1}^hk2^k=\sum_{k=2}^{h+1}(k-1)2^k-\sum_{k=1}^hk2^k\\ =h2^{h+1}+\sum_{k=2}^{h}(k-1)2^k-\sum_{k=1}^hk2^k\\ =h2^{h+1}+\sum_{k=2}^{h}k2^k-\sum_{k=2}^h2^k-\sum_{k=1}^hk2^k\\ =h2^{h+1}-\sum_{k=2}^h2^k-2\\ =h2^{h+1}-(2^{h+1}-4)-2\\ =(h-1)2^{h+1}+2. $$

Martin Argerami
  • 205,756