Determine the value of :

(1+1^-2+2^-2)^1/2 + (1+2^-2+3^-2)^1/2 +......+ (1+2010^-2+2011^-2)^1/2

*** For an extra challenge, solve this puzzle using only pen and paper.

10   for I=1 to 2010
20     Tot=Tot+sqrt(1+1/(I*I)+1/((I+1)*(I+1))):print Tot
30     if I=40 then stop
40   next
50   print Tot

The totals after each of the first 40 terms look like this:

1.5
2.6666666666666666666
3.7499999999999999998
4.7999999999999999998
5.8333333333333333331
6.8571428571428571426
7.8749999999999999997
8.8888888888888888886
9.8999999999999999996
10.9090909090909090905
11.9166666666666666661
12.9230769230769230763
13.9285714285714285708
14.9333333333333333326
15.9374999999999999993
16.9411764705882352934
17.9444444444444444437
18.9473684210526315781
19.9499999999999999991
20.95238095238095238
21.9545454545454545445
22.9565217391304347815
23.9583333333333333322
24.9599999999999999988
25.9615384615384615373
26.9629629629629629617
27.9642857142857142843
28.9655172413793103434
29.9666666666666666651
30.9677419354838709661
31.9687499999999999984
32.9696969696969696953
33.9705882352941176454
34.9714285714285714268
35.9722222222222222204
36.9729729729729729711
37.9736842105263157875
38.9743589743589743569
39.9749999999999999979
40.9756097560975609735

which are 2 - 1/2, 3 - 1/3, 4 - 1/4, etc.

When continued, the ultimate (2010th) total is shown as 2010.9995027349577323635, which checks out as 2011 - 1/2011 and is calculated as:

? 1/2011
 0.0004972650422675285
 
which indeed complements the 2010th sum to a full 2011.

So the answer is 2011 - 1/2011.

 

Posted by Charlie on 2011-12-25 14:42:59