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)
Hello,
You will need this formula : 1+2+...+n=n(n+1)/2
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.
Note that 5=2+3, 8=2+3*2, etc... and 2003=2+3*667b) Simplify: 1/(2*5) + 1/(5*8) + 1/(8*11) + ... + 1/(2000*2003)
Thus the sum is
Let's find a telescoping sum (in most of this type of exercises, it's the trick).
Find a and b such that
Get the common denominator of the RHS :
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
And then
\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}