Given 'x' not equal to 0, let us consider the follwoing relation:
x + x + x + .... +x (added 'x' times) = x²
Differentiating both sides with respect to x, we get:
1 + 1 + 1 + 1 + .... + 1 ('x' times) = 2x
(Since the derivative of x² with respect to 'x' is 2x).
So we now have:
x = 2x
Cancelling 'x' from both sides, we have:
1 = 2
Now the very obvious question follows:
Where is the flaw ?
(In reply to Puzzle Solution
by K Sengupta)
The method as given in the problem would also have failed if the number of times were a variable but the quantity to be summed were a constant.
Thus, if h(x) = (n+n+.....+n) (x times), where n is a constant.
Then apparently, h'(x) = (0+0+....+0) (x times) = 0
which is incorrect as d/dx (nx) = n
However, remembering the requirement of differentiation upon the number of times, we obtain:
= (0+0+....+0) (x times)+ (n+n+....+n) ( 1 time)
= 0*x + n*1
= n, which is indeed true.
Edited on September 26, 2007, 12:52 pm