# Maths Problems

• Jul 24th 2008, 03:41 AM
xwrathbringerx
Maths Problems
a) Is it possible to arrange the integers from 1 to 1000 in a table with 20 rows and 50 columns so that if for each of the rows all the numbers in it are added together, then twenty consecutive integers are obtained?

b) Simplify: 1/(2*5) + 1/(5*8) + 1/(8*11) + ... + 1/(2000*2003)
• Jul 24th 2008, 04:02 AM
kalagota
Quote:

Originally Posted by xwrathbringerx
b) Simplify: 1/(2*5) + 1/(5*8) + 1/(8*11) + ... + 1/(2000*2003)

\sum_{n=1}^{667} \frac{1}{(3n-1)(3n+2)} =
\sum_{n=1}^{667} \frac{1}{9n^2+3n-2}
• Jul 24th 2008, 04:07 AM
Moo
Hello,

You will need this formula : 1+2+...+n=n(n+1)/2
Quote:

Originally Posted by xwrathbringerx
a) Is it possible to arrange the integers from 1 to 1000 in a table with 20 rows and 50 columns so that if for each of the rows all the numbers in it are added together, then twenty consecutive integers are obtained?

Let n, n+1, ..., n+19 be these 20 consecutive integers.
The sum of n, n+1, ..., n+19 is also the sum of the integers from 1 to 1000.

n+(n+1)+...+(n+19)=20n+(1+2+...+19)=20n+20*19/2=20n+190
1+...+1000=1000*1001/2=500500

So find n such that 20n+190=500500
2n=50050-19, which is odd, whereas 2n is obviously even. Thus it is not possible to find an integer n satisfying the conditions you want.

Therefore it is not possible to draw such a table.

Quote:

b) Simplify: 1/(2*5) + 1/(5*8) + 1/(8*11) + ... + 1/(2000*2003)
Note that 5=2+3, 8=2+3*2, etc... and 2003=2+3*667

Thus the sum is http://www.gnux.be/latex/data/b270f1...cb48a74c7e.png
Attachment 7330

Let's find a telescoping sum (in most of this type of exercises, it's the trick).
Find a and b such that http://www.gnux.be/latex/data/f1bef6...1ce343f5fc.png
Attachment 7331

Get the common denominator of the RHS :
http://www.gnux.be/latex/data/4ef3b8...d5310c73b1.png
Attachment 7332

Therefore 3a+3b=0 and 5a+2b=1
--> b=-a --> 5a-2a=1 --> a=1/3 and b=-1/3

Thus this sum is equal to

http://www.gnux.be/latex/data/31e6f6...361994188b.png
Attachment 7333

And then http://www.gnux.be/latex/data/8541b7...ec47636a9f.png
Attachment 7334

\frac{1}{(2+3k)\cdot (2+3(k+1))}=\frac{a}{2+3k}+\frac{b}{2+3(k+1)}

\begin{aligned} \frac{1}{(2+3k) \cdot (2+3(k+1))}=\frac{a}{2+3k}+\frac{b}{2+3(k+1)} &=& \frac{a(2+3(k+1))+b(2+3k)}{(2+3k) \cdot (2+3(k+1))} \\ \\
&=& \frac{2a+3ak+3a+2b+3kb}{(2+3k) \cdot (2+3(k+1))} \\ \\
&=& \frac{k(3a+3b)+5a+2b}{(2+3k) \cdot (2+3(k+1))} \end{aligned}

\frac 13 \sum_{k=0}^{666} \frac{1}{2+3k}-\frac{1}{2+3(k+1)}=\frac 13 \left(\frac 12-\frac 15+\frac 15-\frac 18+\frac 18-\dots-\frac{1}{2000}+\frac{1}{2000}-\frac{1}{2003} \right)

S=\frac 13 \left(\frac 12-\frac{1}{2003}\right)=\frac 13 \left(\frac{2001}{4006}\right)=\frac{667}{4006}