Find a closed form for  $$S_n = 1 \cdot 1! + 2 \cdot 2! + \ldots + n \cdot n!.$$ for integer $n \geq 1.$ Your response should have a factorial.  Another induction problem, I tried some examples but they didn't really tell me anything.
122894
I came across this question in some extracurricular problem sets my professor gave me: what is the closed form notation for the following sum:  $$S_n = 1\cdot1!+2\cdot2!+ ...+n \cdot n!$$  I tried computing some terms, and the only "vague" thing I noticed was that maybe I should be subtracting a term, but I'm really not sure. I went around looking on StackExchange's archives for a closed form of $S_n = 1!+2!+ ...+ n!$ but that didn't help me with my problem much.   Any pointers?
99967
For a positive integer $n$, prove that: $$\sum_{d\mid n} (-1)^{n/d} \cdot \varphi(d) = \begin{cases} 0 &amp; \text{if $n$ is even}  \\ -n  &amp; \text{if $n$ is odd}  \end{cases}$$  I can't even think, how, to begin with, this problem, please provide easy proof of this.
25221
Prove by induction that for all $n \ge 0$:  $${n \choose 0} + {n \choose 1} + ... + {n \choose n} = 2^n.$$  In the inductive step, use Pascal’s identity, which is:  $${n+1 \choose k} = {n \choose k-1} + {n \choose k}.$$  I can only prove it using the binomial theorem, not induction.
81069
Use Mathematical induction to prove that for all integers, $n$ is greater than or equal to $1$. I am confused on what to do after I do the the basis step that is using $n$ as $1$.   $$\frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} +...+\frac{1}{n(n+1)}=1-\frac{1}{n+1}$$
115600
Use induction to prove that that $8^{n} | (4n)!$ for all positive integers $n$  So far I have: Base case (n = 1) = $8^{1} | (4(1))!$  = $8 | 24$ which is true.  Induction Step:  $8^{n + 1} | (4(n + 1))!$  $8^{n + 1} | (4n + 4)!$   A bit confused as to how to close this proof out, also wanted to make sure my current progress is correct as well. Any help is appreciated.
110000
How can I prove that $n!&lt;n^n$ for every $n&gt;1$ using the induction principle?  If I put $n=2$ I get $2&lt;4$ so I know that $n!&lt;n^n$ is true.  Now I don't know how to prove that $(n+1)!&lt;(n+1)^{n+1}$. I can write $(n+1)!=n!(n+1)$ but then I don't know how to go on.
114867
How to get a closed form for $$\sum_{k=0}^{n-p}\binom{n}{k}\binom{n}{p+k}\;?$$  I tried to write binominal in term of gamma function but I got no result. What is your suggestion to solve the problem?
19547
Prove by Mathematical Induction . . .  $1(1!) + 2(2!) + \cdot \cdot \cdot +n(n!) = (n+1)!-1$  I tried solving it, but I got stuck near the end . . .  a. Basis Step:      $(1)(1!) = (1+1)!-1$      $1 = (2\cdot1)-1$      $1 = 1 \checkmark$   b. Inductive Hypothesis     $1(1!) + 2(2!) + \cdot \cdot \cdot +k(k!) = (k+1)!-1$   Prove k+1 is true.     $1(1!) + 2(2!) + \cdot \cdot \cdot +(k+1)(k+1)! = (k+2)!-1$      $\big[RHS\big]$      $(k+2)!-1 = (k+2)(k+1)k!-1$      $\big[LHS\big]$      $=\underbrace{1(1!) + 2(2!) + \cdot \cdot \cdot + k(k+1)!} + (k+1)(k+1)!$ (Explicit Last Step)      $= \underbrace{(k+1)!-1}+(k+1)(k+1)!$ (Inductive Hypothesis Substitution)      $= (k+1)!-1 + (k+1)(k+1)k!$      $= (k+1)k!-1 + (k+1)^{2}k!$   My [LHS] looks nothing like my [RHS] did I do something wrong?  EDIT:     $ = (k+1)k! + (k+1)^2k! -1 $      $ = (k+1)(k!)(1 + (k+1))-1$      $ = (k+1)(k!)(k+2)-1 = (k+2)(k+1)k!-1$
90457
I have a hard induction problem here that I feel completly lost on.  For every $n\ge3$ there is n different positive integer $a_1,...,a_n$ so that $\frac{1}{a_1}+\frac{1}{a_2}+...+\frac{1}{a_n}=1$.  There is only one possible solution for $n=3 $  $\frac{1}{2}+\frac{1}{3}+\frac{1}{6}=1$  for $n=4$ there is 2 possible solutions and they both start with $\frac{1}{2}$.  $\frac{1}{2}+\frac{1}{3}+\frac{1}{8}+\frac{1}{24}$ and $\frac{1}{2}+\frac{1}{4}+\frac{1}{5}+\frac{1}{20}$.  I think the first one is the one.  For $n=5$ there is alot of different possible solutions but it have to start with at least $\frac 13$  I think it have to start with $\frac 12+\frac13$ and then be some kind of sum answer for the last $\frac16$ based on $n$. But I don't know  I have tried finding a pattern but not getting anywhere anymore. Any tips on how to tackle a problem like this?
59739
How to find the tight bound for the binomial series $\binom{n}{2}\cdot2^2+\binom{n}{4}\cdot2^4+\binom{n}{6}\cdot2^6\ldots,+ \binom{n}{n}\cdot2^n$ I have found $3^n$ for series $(1+2)^n=1+\binom{n}{1}\cdot2^1+\binom{n}{2}\cdot2^2+\binom{n}{3}\cdot2^3+\cdots+ \binom{n}{n}\cdot2^n$ Is there any way to prove the bound of above series.
73531
If you expand $(x_1+x_2+\cdots+x_k)^n$, how many terms will there be once you collect terms with equal monomials? What is the sum of all the coefficients?  I'm kind of lost here. This came up with other questions on the topic of binomial coefficients, and i have figured those out. I have spent almost an hour trying to figure where to start and haven't made any progress.
72210
I am trying to prove that $\dfrac{(2n)!}{2^n}$  is integer. So I have tried it by induction, I have took  $n=1$, for which we would have $2/2=1$ is integer. So for $n=k$ it is true, so now comes time to proof it for $k+1$, $(2(n+1))!=(2n+2)!$, which is equal to $$1 \times 2 \times 3 \times \cdots \times (2n) \times (2n+1) \times (2n+2),$$ second term would be $$2^{n+1}=2 \times 2^n$$ Finally if we divide $(1 \times 2 \times 3 \times \cdots \times (2n) \times (2n+1) \times (2n+2)$ by $2^{n+1}=2 \times 2^n$,and consider that,$(2n)!/(2^n)$  is integer, we get $(2n+1) \times (2n+2)/2=(2n+1) \times 2 \times (n+1)/2$, we can cancel out $2$, we get $(2n+1)(n+1)$ which is definitely integer. I am curious this so simple? Does it means that I have proved correctly?
28191
I'm sure there's an identity or some known "trick" to solve this but I couldn't find it (Looked in Wikipedia, Wolfram and here):  $$ \sum_{k=0}^n (-1)^k{n \choose k} =0 $$  I want to know how this can be proven. I tried induction but I'm not sure what to do with ${{n+1}\choose k}$.  Thanks!
115930
Question :   Determine all natural numbers n such that 3   divides $n\cdot2^{n}+1$  Actually I don't have any ideas to approach but  my efforts :   I see $n=1,2,7,8,13,14$ so I think :   $n=6k+1$ and $n=6k+2$ $k\in \Bbb{N} $  If I'm not wrong but I don't know how ? I prove it ?
215
Use induction to prove the following: $1! + 2! + .... + n! &lt; (n + 1)!$  Base case:  $n = 1$  $1! &lt; 2!$ true   Inductive step: Assume that $1! + 2! + .... + k! \le  (k + 1)!$ is true   let $n = k + 1$  $1! + 2! + .... + (k + 1)! &lt; (k + 2)!$  $1! + 2! + .... + k! + (k + 1)! &lt; (k + 2)!$  $1! + 2! + .... + k! + (k + 1)! &lt; (k + 2)!$  $(k + 1)! + (k + 1)! &lt; (k + 2)!$  $2(k + 1)! &lt; (k + 2)!$  $2(k + 1)(k)(k - 1).. &lt; (k + 2)(k + 1)(k)(k - 1)....$  $2 &lt; (k + 2)$   the right is always bigger for k are non negative integer.
21948
Given $$1\cdot 2^1 + 2\cdot 2^2 + 3\cdot 2^3 + 4\cdot 2^4 + \cdots + d \cdot 2^d = \sum_{r=1}^d r \cdot 2^r,$$ how can we infer to the following solution? $$2 (d-1) \cdot 2^d + 2. $$   Thank you
96023
Is there a formula for the following sum?  $S_n = 1\cdot2 + 2\cdot 3 + 3\cdot 4 + 4\cdot 5 +\ldots  + n\cdot (n+1)$