When I'm reading Computer Systems: A Programmer's Perspective, I met the sum of binary numbers and failed to prove it:  $$ 2^0 + 2^1 + 2^2 + 2^3 + \cdots + 2^n = 2^{n+1} - 1 $$  This might be preliminary knowledge, I'm not good at mathematics, any body could give me a hint?
45648
Since everyone freaked out, I made the variables are the same. $$ \sum_{x=1}^{n} 2^{x-1} $$  I've been trying to find this for a while. I tried the usually geometric equation  but I couldn't get it right (if you need me to post my work I will). Here's the outputs I need:  1, 3, 7, 15, 31, 63  If my math is correct.
106527
I believe I see that $a_n = 2^n(a_0+1) - 1$ but am somewhat uncertain where to proceed afterwards. I am a complete beginner at number theory and would appreciate it if someone could point me in the right direction--surely there is some obvious argument I am missing.
49569
How to find the sum of this series?  $$\sum_{k=0}^{\infty}\cfrac{{2}^{k}}{\binom{2k+1}{k}}$$  It seems very easy. But I still can not work it out, can anyone help?
81866
Since last week I have been working on a way, how to sum $1$ and $2$ to have $133$.  So for instance we can have $133$ $1s$ or $61$ $s$2 and one and so on. Looking back to the example: if we sum: $1 + 1... + 1 = 133$ there is only one way. But for the second one there will be $131$ possible ways. And I have to do this for every possible combination. I am stuck with it and I have no idea whatsoever on how to begin.  Any ideas or methods I could use guys?
82684
I've worked out a proof, but I was wondering about alternate, possibly more elegant ways to prove the statement. This is my (hopefully correct) proof:  Starting from the identity $2^m = \sum_{k=0}^m \binom{m}{k}$ (easily derived from the binomial theorem), with $m = 2n$:  $2^{2n} = 4^n = \binom{2n}{0} + \binom{2n}{1} + \cdots + \binom{2n}{2n-1} + \binom{2n}{2n}$  Applying the property $\binom{m}{k} = \binom{m}{m-k}$ to the second half of the list of summands in RHS above:  $4^n = \binom{2n}{0} + \binom{2n}{1} + \cdots + \binom{2n}{n-1} + \binom{2n}{n} + \underbrace{\binom{2n}{n-1} +\cdots \binom{2n}{1} + \binom{2n}{0}}_{\binom{m}{k} = \binom{m}{m-k} \text{ has been applied}}$  Rearranging the above sum by alternately taking terms from the front and end of the summand list in RHS above (and introducing the term $\binom{2n}{-1} = 0$ at the beginning just to make explicit the pattern being developed):  $4^n = (\binom{2n}{-1} + \binom{2n}{0}) + (\binom{2n}{0} + \binom{2n}{1}) + \cdots + (\binom{2n}{n-1} + \binom{2n}{n})$  Finally, using the property $\binom{m}{k} + \binom{m}{k-1} = \binom{m+1}{k}$ on the paired summands, we get the desired result:  $4^n = \binom{2n+1}{0} + \binom{2n+1}{1} + \cdots + \binom{2n+1}{n}$
50457
How can I show that the sum   $$ \sum_{k = 0}^n 2^k \binom{n}{k}$$  is equal to $3^n$?
7233
How can I prove the identity:  $$ \sum_{k=0}^n\binom{2n}{2k}^2-\sum_{k=0}^{n-1}\binom{2n}{2k+1}^2=(-1)^n\binom{2n}{n}? $$  Maybe, can we expand $$ f(x)=(1+x)^{2n}? $$  Thank you.
119270
$$\sum_{i=1}^n i^2 C^i_n = n(n+1) \cdot 2^{n-2}$$ I could use some help with proving that this equation stands. Should I try to use mathematical induction?
100546
Everyone knows about the classic  $$ \sum_{i=1}^{\infty} \dfrac{1}{2^i} = 1 $$ However, is there any way to find $$ \sum_{i=0}^{\infty} \dfrac{1}{2^{2^i}} = \dfrac12 + \dfrac14 + \dfrac{1}{16} + \dfrac{1}{256} + \cdots $$
10232
I came across a question where I needed to find the sum of the factorials of the first $n$ numbers. So I was wondering if there is any generic formula for this? Like there is a generic formula for the series: $$ 1 + 2 + 3 + 4 + \cdots + n = \frac{n(n+1)}{2} $$ or $$ 1^{2} + 2^{2} + 3^{2} + 4^{2} + \cdots + n^{2} = \frac{n(n+1)(2n + 1)}{6} $$  Is there is any formula for: $$ 1! +2! +3! + 4! + \cdots + n! $$ and $$ {1!}^2 +{2!}^2 +{3!}^2 + \cdots + {n!}^2 $$?  Thanks in advance. If not, is there any research on making this type of formula? Because I am interested.
83623
I'm trying to prove the following equation above. So far I have: \begin{align} 2^{2n} &amp;= (1+1)^{2n}\\ &amp;= \sum_{k=0}^{2n}\binom{2n}{k}1^k1^{n-k} = \sum_{k=0}^{2n}\binom{2n}{k} &amp; \text{(By the Binomial Theorem)} \end{align}  I know I have to use the following identity somehow: $$\binom{n+1}{k} = \binom{n}{k-1} + \binom{n}{k}$$  How do I split my summation to get what I'm looking for? Thanks!  EDIT: HERE IS MY SOLUTION \begin{align*} 2^{2n} &amp;= (1+1)^{2n}\\ &amp;= \sum_{k=0}^{2n}\binom{2n}{k}1^k1^{2n-k} &amp; \text{(By the Binomial Theorem)}\\ &amp;= \sum_{k=0}^{n}\binom{2n}{k} + \sum_{k=n+1}^{2n}\binom{2n}{k}\\ &amp;= \sum_{k=0}^{n}\binom{2n}{k} + \sum_{k=n+1}^{2n}\binom{2n}{2n-k} &amp; \text{(Binomial Symmetry)}\\ &amp;= \sum_{k=0}^{n}\binom{2n}{k} + \sum_{k=0}^{n-1}\binom{2n}{k}\\ &amp;= \sum_{k=0}^{n}\binom{2n}{k} + \sum_{k=1}^{n}\binom{2n}{k-1}\\ &amp;= \binom{2n}{0} + \sum_{k=1}^{n}\binom{2n}{k} + \sum_{k=1}^{n}\binom{2n}{k-1}\\ &amp;= \binom{2n}{0} + \sum_{k=1}^{n}\binom{2n+1}{k} &amp; \text{(By Identity listed above)}\\ &amp;= \sum_{k=0}^{n}\binom{2n+1}{k} \end{align*}
69136
Give a combinatorial proof of the following identity:  $$3^n=\sum_{i=0}^{n}\binom{n}{i}2^{n-i}$$  I can't see any counting argument that would yield $3^n$, and the right hand side is also pretty opaque. For some reason I really really suck at doing these proofs - I just started my first combinatorics course.
14550
Problem:  Let $x, y,$ and $z$ be three integers. Suppose that the binary representation of $x$ uses $n$ bits, the binary representation of $y$ uses $n$ bits and the binary representation of $z$ uses $n$ bits. How many bits does the binary representation of the product $xyz$ use?  Thoughts:  My first impulse is to consider the case where $x,y,z$ are all the maximal number that can be represented by $n$ bits in binary, namely that $x=y=z=2^n -1 $.   Then $xyz = (2^n -1)^3 = 2^{3n}-2^{2n}-2^{2n+1}+2^{n+1}+2^n-1$  $=2^{2n}(2^n-1)-2^{n+1}(2^n-1)+2^n-1 = (2^{n}-1)(2^n-2^{n+1}+1)$  Expanding in this manner has not really helped me clarify what the solution should be. Any different approaches or hints would be much appreciated.
39049
Given the identity  $$\sum_{i=0}^{n} \binom{k-1+i}{k-1} = \binom{n+k}{k}$$  Need to give a combinatorial proof   a) in terms of subsets  b) by interpreting the parts in terms of compositions of integers  I should not use induction or any other ways...  Please help.
31421
Why should we expect that $$2^n=\sum_{k=0}^n {n\choose k}$$   It is easily seen to be true, by the binomial theorem:  just set $x=y=1$ in $(x+y)^n$.  But what is an intuitive reason why it is true (in terms of subsets)?
25606
Can someone give me a hint how to evaluate $$ \binom{n}{0}+\binom{n}{2}+\cdots+\binom{n}{o(n)},$$ where $o(n)$ is $n$ if $n$ is even and $n-1$ otherwise ?
7156
Prove that   $$\sum^{n}_{i=1}\binom{n}{i}i=n2^{n-1}$$   I can't find counting interpretations for either of the sides. A hint of "if $S$ is a subset of $\{1, . . . , n\}$ and $S^\prime$ is its complement then $|S| + |S^\prime| = n$" was also given, but I still don't know how to begin.