Hello. I have some sequences. I need to show if they are correct and what is it's limit when it is infinite.

Question

osanseviero · Answer

$$1+3+5+7+...+(2n-1)=n ^{2}$$
I understand everything except the n*n at the end. Why is that?

anonymous · Answer

$$S=1+3+5+\cdots+(2n-3)+(2n-2)+(2n-1)\
S=(2n-1)+(2n-2)+(2n-3)+\cdots+5+3+1$$
Adding vertically, you have
$$\large2S=\underbrace{2n + 2n + 2n+\cdot+2n+2n+2n}_{n\text{ terms}}\
S=n+\cdots +n\
S=n\cdot n\
S=n^2$$

osanseviero · Answer

thanks a lot :D

osanseviero · Answer

$$2 + 3 + 5 + 8 + ... + (3n-1)=n(3n+1)\div2$$
That is not correct, isnt it?

osanseviero · Answer

if it isnt true, can there be a limit? (infinite in this one)

anonymous · Answer

I think there's something wrong with the given terms. If that's supposed to be the sum of multiples of 3 minus 1, then the sequence is
$$2+5+8+\cdots+(3n-3)+(3n-2)+(3n-1)$$
Then, using similar reasoning as for the first one,
$$2S=n(3n+1)$$
So yes, you're right.

osanseviero · Answer

Thanks a lot, the question asked me to see if the sequence was correct, and say the limit...I have a bunch of this to do :)