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Does this work?(or did i do it wrong?)fp1 proof by induction

∑r=1nr(r+1)=n(n+1)(n+2)3\displaystyle\sum_{r=1}^n r(r+1) = \dfrac{n(n+1)(n+2)}{3}


n=1
∑r=11r(r+1)=1(1+1)(1+2)3\displaystyle\sum_{r=1}^1 r(r+1) = \dfrac{1(1+1)(1+2)}{3}


2=22=2


n=k
∑r=1kr(r+1)=k(k+1)(k+2)3\displaystyle\sum_{r=1}^k r(r+1) = \dfrac{k(k+1)(k+2)}{3}



n=k+1
∑r=1k+1r(r+1)=∑r=1k+1r(r+1)+[k+1(k+1+1)]\displaystyle\sum_{r=1}^{k+1} r(r+1) = \sum_{r=1}^{k+1} r(r+1) +[k+1(k+1+1)]



∑r=1k+1r(r+1)=k(k+1)(k+2)3+[k+1(k+2)]\displaystyle\sum_{r=1}^{k+1} r(r+1) = \dfrac{k(k+1)(k+2)}{3}+[k+1(k+2)]



∑r=1k+1r(r+1)=k(k+1)(k+2)+3[(k+1)(k+2)]3\displaystyle\sum_{r=1}^{k+1} r(r+1) = \dfrac{k(k+1)(k+2)+3[(k+1)(k+2)]}{3}



∑r=1k+1r(r+1)=(k+3)(k+1)(k+2)3\displaystyle\sum_{r=1}^{k+1} r(r+1) = \dfrac{(k+3)(k+1)(k+2)}{3}



∑r=1k+1r(r+1)=k+1(k+1+1)(k+1+2)3\displaystyle\sum_{r=1}^{k+1} r(r+1) = \dfrac{k+1(k+1+1)(k+1+2)}{3}


since true for n=1 it's therefore true for n=2,3,4.....


also what does nez+nez^+ mean? it's looks really cool but i don't understand it
(edited 7 years ago)
Yes, it looks right, watch out for bracketing errors.

This:

n=k+1


Should be this:

[br]∑r=1k+1r(r+1)=∑r=1kr(r+1)+(k+1)(k+1+1)[br] \displaystyle\sum_{r=1}^{k+1} r(r+1) = \sum_{r=1}^{k} r(r+1) +( k+1)(k+1+1)

But other than that it's all right.
Original post by huiop



also what does nez+nez^+ mean? it's looks really cool but i don't understand it


n∈Z+ n \in \mathbb Z_+ means " n is an element of the set of positive integers"
Original post by huiop
Proof by induction...


Good (though that 2=2 is pointless). Just be careful of the brackets; I'm sure you got them in there correctly on the paper.

n∈Z+n\in \mathbb{Z_+} means nn belongs to the set of positive integers. Another indication could be n∈Nn\in \mathbb{N} where it means that nn is in the subset of natural numbers, n≠0n\not=0
(edited 7 years ago)
You must not write 2 = 2 that is not correct.

Write LHS = 2 RHS = 2 so true for n = 1.
Original post by huiop


Your proof needs to have some explanation too - also see my comment above.
For example,
'Assume true for some n = k'
Reply 6
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huiop
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Original post by NotNotBatman
Yes, it looks right, watch out for bracketing errors.

This:

n=k+1


Should be this:

[br]∑r=1k+1r(r+1)=∑r=1kr(r+1)+(k+1)(k+1+1)[br] \displaystyle\sum_{r=1}^{k+1} r(r+1) = \sum_{r=1}^{k} r(r+1) +( k+1)(k+1+1)

But other than that it's all right.

oops, yes i was trying to correct them all before the replies came :smile:
Original post by NotNotBatman
n∈Z+ n \in \mathbb Z_+ means " n is an element of the set of positive integers"

???? i don't quite understand
Original post by RDKGames
Good (though that 2=2 is pointless). Just be careful of the brackets; I'm sure you got them in there correctly on the paper.

n∈Z+n\in \mathbb{Z_+} means nn belongs to the set of positive integers. Another indication could be n∈Nn\in \mathbb{N} where it means that nn is in the subset of natural numbers, n≠0n\not=0

oops yea... corrected now...

so is it suppsoed to mean n is any positive number? i don't quite understand i know n by itself just means n, the e stands for any then the z i don't know what that is :/
Original post by Muttley79
You must not write 2 = 2 that is not correct.

Write LHS = 2 RHS = 2 so true for n = 1.

o.o
ok so that's a better way to express it?
Original post by Muttley79
Your proof needs to have some explanation too - also see my comment above.
For example,
'Assume true for some n = k'


ok thanks
Original post by huiop

oops yea... corrected now...

so is it suppsoed to mean n is any positive number? i don't quite understand i know n by itself just means n, the e stands for any then the z i don't know what that is :/


Top line of: http://www.math.ku.edu/~porter/Math_symbols%20.pdf

By saying n∈Zn \in \mathbb{Z} you are restricting what types of values n can be. The Z gets rid off any fractions and irrationals, thus leaving positive/negative integers. By saying n∈Z+n \in \mathbb{Z_+} you are specifying it further down to only POSITIVE integers.
Reply 8
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huiop
OP
1
Original post by RDKGames
Top line of: http://www.math.ku.edu/~porter/Math_symbols%20.pdf

By saying n∈Zn \in \mathbb{Z} you are restricting what types of values n can be. The Z gets rid off any fractions and irrationals, thus leaving positive/negative integers. By saying n∈Z+n \in \mathbb{Z_+} you are specifying it further down to only POSITIVE integers.


ooh ok so it means any positive integer thanks a ton

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