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0 votes
1 answer
37 views

Checking an induction proof for a summation.

In my textbook practice problem, I want to prove the following by induction: $$\sum_{i=0}^{n}\sum_{k=0}^{n-i}a_{i, k} = \sum_{m=0}^{n}\sum_{k=0}^{m}a_{m-k, k}$$ For my "$n+1$ implies $n$" ...
Newbie1000's user avatar
0 votes
1 answer
23 views

Evaluation of an indexed sum. [closed]

I am going through a textbook solution to an induction problem and I'm not sure why it says this summation holds. $\sum_{m=0}^0 \sum_{k=0}^m a_{m - k, k} = a_{0, 0}$ This is very simple I know, but ...
Newbie1000's user avatar
3 votes
1 answer
66 views

How can i simplify the following formula: $\sum\limits_{i,j=1}^{n}(t_{j}\land t_{i})$?

Consider the following time discretization $t_{0}=0< t_{1} < ... < t_{n} = T$ of $[0,T]$ where the time increments are equal in magnitude, i.e. $t_{j}-t_{j-1}=\delta$. How can i simplify the ...
SABOY's user avatar
  • 1,838
2 votes
3 answers
92 views

Proof $\sum_{k=1}^{2n} (-1)^{1+k}\frac{1}{k} = \sum_{k=1}^{n}\frac{1}{n+k}$ by induction. [duplicate]

I'm trying to show the following formula: $$\sum_{k=1}^{2n} (-1)^{1+k}\frac{1}{k} = \sum_{k=1}^{n}\frac{1}{n+k}$$ I have already verified the formula with $n=1$, now I continue with the induction: $$ \...
Ivan Camilo Ballen Mendez's user avatar
1 vote
2 answers
52 views

Have I followed through on this proof by induction correctly?

We are asked to prove that $1^3+2^3+\cdots+n^3 = (1+2+\cdots + n)^2$ by induction. Basis: $n=1 \\ 1^3 = 1^2 \quad\checkmark \\ n=2 \\ 1^3+2^3 = (1+2)^2 = 9 \quad\checkmark $ We rewrite both series ...
David Farmilant's user avatar
1 vote
3 answers
88 views

Bounded sequence of reals

Suppose that $\{a_{n} \}$ is a sequence of real numbers that satisfy $a_{i} + a_{j} \geq a_{i+j}$. Then prove $$a_{n} \leq \sum_{i=1}^{n} \frac{a_{i}}{i}$$. I tried to use straight-up induction, but ...
user100101212's user avatar
1 vote
6 answers
143 views

Prove that $\sum_{k=0}^{n}{\frac{1}{k!}}\leq 3$ $\forall n\in \mathbb{N}$

Prove that $\sum_{k=0}^{n}{\frac{1}{k!}}\leq 3$ $\forall n\in \mathbb{N}$ I proved it by induction: "$n=1$" $ \quad \sum_{k=0}^{1}{\frac{1}{k!}}\leq 3 \quad \checkmark$ "$n \implies n+1$": $$\...
Analysis's user avatar
  • 2,482
1 vote
1 answer
123 views

Suppose $(a)_{j=1}^{\infty}$ is a sequence of real numbers. prove by induction on n that $|\sum_{j=1}^{n}a_j|\leq\sum_{j=1}^{n}|a_j|$

This is a proof that my teacher gave I'm having a hard time with the last line of the proof. Suppose $(a)_{j=1}^{\infty}$ is a sequence of real numbers. prove by induction on n that $$|\sum_{j=1}^{n}...
jack's user avatar
  • 513
0 votes
3 answers
91 views

Prove through induction that $\sum_{a=1}^{b}a(b-a)=\frac{b(b-1)(b+6)}{6}$ [closed]

Given that $\sum_{a=1}^{b}a=\frac{a(a+1)}{6}$ prove through induction that $$\sum_{a=1}^{b}a(b-a)=\frac{b(b-1)(b+6)}{6}$$ Normally I would start by showing that this statement is true for $b=1$ and ...
Ski Mask's user avatar
  • 1,928
-1 votes
3 answers
112 views

What does $\sum_{k=1}^{n}k=\frac{n(n+1)}{2}$ mean in induction? [closed]

If you're given that $$\sum_{k=1}^{n}k=\frac{n(n+1)}{2}$$ and let's say that prove that $\sum_{k=1}^{n}k+\frac{1}{k+n}=blah$, $n>0$ what does $k$ in itself equal? I mean if you had to put $k$ ...
Ski Mask's user avatar
  • 1,928
4 votes
2 answers
425 views

How to prove this summation formula?

Mathematica is able to calculate the following sum: \begin{align} &\sum_{k,l=0}^{n,m}\frac{(-1)^{n+m-l-k}(2m)!(2n+1)!(2 )^{2k+2l}}{(2k+1)!(2l)!(n-k)!(m-l)!}(k+l)! \nonumber\\ =&\frac{(-1)^{...
Gesbesgue's user avatar
  • 147
0 votes
3 answers
109 views

Mathematical induction proof problem [closed]

How to prove with mathematical induction that $$\sum _{k=1}^{n}\frac{2k-1}{2^k}=3-\frac{2n+3}{2^n}$$ if $n \in \mathbb N$?
Mari's user avatar
  • 9
4 votes
3 answers
384 views

Induction on inequalities: $\frac1{1^2}+\frac1{2^2}+\frac1{3^2}+\ldots+\frac1{n^2}<2$ [duplicate]

I am trying to solve this inequality by induction. I just started to learn induction this week and all the inequalities we had been solved were like an equation less than another equation (e.g. $n! \...
SC L's user avatar
  • 45
2 votes
1 answer
4k views

The identity $a^n-b^n=(a-b) (\sum_{i=0}^{n-1}a^ib^{n-1-i})$ [duplicate]

How do I use finite induction to prove that $$a^n-b^n=(a-b) (\sum_{i=0}^{n-1}a^ib^{n-1-i}), \forall a,b\in \Bbb{R}\space \text{and} \space \forall n \in \Bbb{N}?$$ Ok, for $n=2$ it's fine. $a^2-b^2=(a-...
Derso's user avatar
  • 2,773
5 votes
3 answers
4k views

Proof of Equation by Well Ordering Principle

I have an assignment question Prove by either the Well Ordering Principle or induction that for all nonnegative integers $n$: $$\sum_{k=0}^n k^3 = \left(\frac{n(n+1)}{2}\right)^2.$$ I am able to ...
Ankit's user avatar
  • 93

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