Linked Questions

4
votes
7answers
16k views

Evaluate and prove by induction: $\sum k{n\choose k},\sum \frac{1}{k(k+1)}$ [duplicate]

$\displaystyle 0\cdot \binom{n}{0} + 1\cdot \binom{n}{1} + 2\binom{n}{2}+\cdots+(n-1)\cdot \binom{n}{n-1}+n\cdot \binom{n}{n}$ $\displaystyle\frac{1}{1\cdot 2} + \frac{1}{2\cdot 3}+\frac{1}{3\cdot ...
4
votes
6answers
425 views

Proof of equation with binomial coefficients: $\sum\limits_{k=1}^{n} (k+1) \binom{n}{k} = 2^{n-1} \cdot (n+2)-1$ [duplicate]

$$\sum\limits_{k=1}^{n} (k+1) \binom{n}{k} = 2^{n-1} \cdot (n+2)-1$$ Maybe it's simple to prove this equation but I'm not sure how to get along with the induction. Any hints for this? Or may I use ...
4
votes
4answers
1k views

Trying to find $\sum\limits_{k=0}^n k \binom{n}{k}$ [duplicate]

Possible Duplicate: How to prove this binomial identity $\sum_{r=0}^n {r {n \choose r}} = n2^{n-1}$? $$\begin{align} &\sum_{k=0}^n k \binom{n}{k} =\\ &\sum_{k=0}^n k \frac{n!}{k!(n-k)!} =...
0
votes
3answers
1k views

Does $ \sum_{k=0}^n k {n \choose k}$ have a convenient exponential equivalent? [duplicate]

I know that: $${n \choose 0} + {n \choose 1} + ... + {n \choose n} = 2^n.$$ Does $$0 {n \choose 0} + 1 {n \choose 1} + ... + n {n \choose n} = ??$$ have some convenient simplification?
1
vote
5answers
332 views

how to prove: $\sum\limits_{k=1}^n k\binom{n}{k}=n \cdot 2^{n-1} $ [duplicate]

need help to prove this: $\sum\limits_{k=1}^n k\binom{n}{k}=n \cdot 2^{n-1} $ where $n$ is integer $\geq 1$. Question also said taking the derivative of $(1 + x)^n$ would be helpful which I've found ...
3
votes
5answers
223 views

What is the sum of the series with binomial sequences: $\sum_{k=0}^{n} k \binom{n}{k}$? [duplicate]

compute this sum: $\sum_{k=0}^{n} k \binom{n}{k}$ I tried but I got stuck
0
votes
2answers
758 views

How do I prove this using mathematical induction: $\sum_{k = 1}^{n}k\binom{n}{k}=n2^{n-1}$? [duplicate]

$\sum_{k = 1}^{n}k\binom{n}{k}=n2^{n-1}$ How do I prove this using mathematical induction?
0
votes
2answers
1k views

Binomial coefficient identity $\sum_{k=1}^n k {n \choose k } = n\cdot 2^{n-1}$ [duplicate]

I'm having a bit of problems proving the following: $$\sum_{k=1}^n k {n \choose k } = n\cdot 2^{n-1}$$ I always seem to get to the line: $2^{n-1} + 1 = 2^n$ which I know is untrue. Could anyone ...
1
vote
2answers
511 views

The binomial formula. how to show: $\Sigma_{k=0}^n k \binom{n}{k} = n2^{n-1}$ [duplicate]

Does anyone know how to show that: $\Sigma_{k=0}^n k \binom{n}{k} = n2^{n-1}$? I think we are suppose to use the binomial formula for that.. Thank you!
1
vote
2answers
258 views

Prove that $\sum_{k=1}^{n}k\binom{n}{k}=n\cdot2^{n-1}$ [duplicate]

I want to prove the following: $$\sum_{k=1}^{n}k\binom{n}{k}=n\cdot2^{n-1}$$ what I did is(use binominal): $$\binom{n}{k}X^k\cdot 1^{n-k} = (X+1)^n$$ $$k\binom{n}{k}X^k\cdot 1^{n-k} = k(X+1)^k-1$$ now ...
1
vote
4answers
115 views

How to calculate $\sum\limits_{k=0}^{n}{k\dbinom{n}{k}}$ [duplicate]

I derived this sum from a problem I have been working on. Somehow I don't know how to proceed. I only know some basics like $\sum\limits_{k=0}^{n}\dbinom{n}{k} = 2^n$. Meanwhile I am reading the ...
0
votes
1answer
147 views

Sum of weighted binomial coefficients [duplicate]

I am struggling with computing the following sums: $$\sum_{k=1}^{n}k\binom{n}{k}=\binom{n}{1}+2\binom{n}{2}+...+n\binom{n}{n}$$ and $$\sum_{k=0}^{n}\frac{1}{k+1}\binom{n}{k}=\binom{n}{0}+\frac{...
0
votes
3answers
60 views

I have a algorithm and from the post its time complexity is: $\sum_{k=1}^{n}k\binom nk= n 2^{n-1}$ [duplicate]

I have a algorithm and from the post its time complexity is: $$\sum_{k=1}^{n}k\binom nk= n 2^{n-1}$$ My question is, how to get the result $n2^{n-1}$ need some help . thanks
-1
votes
2answers
60 views

Evaluate ${n \choose 1}+2{n \choose 2}+…+n{n \choose n}$ [duplicate]

I have trouble finding an explicit expression to prove this sum using induction, and would like a hint. It would also help if someone provided a non-induction question.
1
vote
3answers
42 views

Show ($n+1$)$2^n$ = $\sum_{i\geq 0}^{} {n + 1\choose i}i$ algebraically. [duplicate]

Show ($n+1$)$2^n$ = $\sum_{i\geq 0}^{} {n + 1\choose i}i$ algebraically. I know $2^n$ = $\sum_{i\geq 0}^{} {n\choose i}$. But how do I manipulate the $(n+1)$ to make it look like the right side?

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