Answer Simple Question

[Ragib]

Ok Guys, you have to admit, these forums are pretty quiet for some reason. Im going to post a problem for someone to figure, ill start off relatively easy:

 

[math](\sum_{n=0}^{m} 10^n)^2[/math] Gives what general Form? Its ok if you can spot the pattern, but try to prove it as well.

[Avisek Roy]

I am not sure or anything, but is this the whole square of a geometric progression, with first term = 10?

[Bluenoise]

Isn't it something like 1010101010101010101010......101010.0

With the number of digits in front of the decimal point being being (m x 2) + 1

 

Sorry been a really long time since I took math.

 

ie.

 

sum of

n = 0 1

n = 1 100

n = 2 10000

n = 3 1000000

.....

n = m 1 x 10^(m x 2)

[Dave]

I am not sure or anything, but is this the whole square of a geometric progression, with first term = 10?

 

Actually, since the summation starts from n = 0, the first term will be 10⁰ = 1. Try to work out an expression for the summation, then square it.

[Ragib]

Lets say m=0, then the Series is equal to 1. m=1, the series is 1 + 10, 11. m=2, 111. Find the pattern in the squares.

[Bluenoise]

opppsss I miss placed that first bracket lol.

 

so it's just 1245678901234567890.....09876543210987654321

 

Which comes from

 

(1 + 10 + 100 + ... + 10^m)(1 + 10 + 100 + ...+ 10^m)

 

=(1 + 10 + 100 + ...+ 10^m) + 10(1 + 10 + 100 + ...+ 100^m) + 100(1 + 10 + 100 + ...+ 10^m) + ... + 10^m(1 + 10 + 100 + ...+ 10^m)

 

111

+ 1110

+ 11100 = 12321

 

Any it gives a sumation like that above with the number of ones in the first term being equal to m and the number of terms being m.

But it's getting late so I got to goto sleep now

[Ragib]

Yes, that is correct. Anyone want to attempt an algebraic proof?

[Bluenoise]

this comes from

 

(1 + 10 + 100 + ... + 10^m)(1 + 10 + 100 + ...+ 10^m)

 

=(1 + 10 + 100 + ...+ 10^m) + 10(1 + 10 + 100 + ...+ 100^m) + 100(1 + 10 + 100 + ...+ 10^m) + ... + 10^m(1 + 10 + 100 + ...+ 10^m)

 

111

+ 1110

+ 11100 = 12321

 

Any it gives a sumation like that above with the number of ones in the first term being equal to m and the number of terms being m.

But it's getting late so I got to goto sleep now

 

repost of my edit.

 

It's a bit of a cop-out at the end but I really should goto sleep lol

[Ragib]

Lol Seeing as your other posts was 9 hours ago, and thinking it was in the afternoon, maybe you should lol.

[Bluenoise]

Lol Seeing as your other posts was 9 hours ago, and thinking it was in the afternoon, maybe you should lol.

 

I just want to point out that you're on the other side of the world from me.....

So it was only 1am here.

[Ragib]

O yes I knew you were in Canada, I just assumed you logged the forums in the afternoon, and 9 hours after that would be pretty late. 1am is quite late, though since its holidays I regularly go to sleep at 3am and wake up at 2pm :S.

 

So to the question, someone try to do it please.

 

Possibly Induction? To tell you the truth guys I havent done it either, but I remember the proof was relatively simple.

 

Heres another short problem, that requires very little knowledge, pretty much just basic algebra.

 

What is the square root of i?

 

Background info for newbies: i is the square root of -1. All complex numbers can be written as a+bi, where a is the real, normal everyday number, and bi and a real everyday number times i. Simple from there.

[Ragib]

Anyone? Come on, at least the 2nd problem...