1@xyz
The simplest method is mathematical induction. Let P(n) be the the conjectured identity. Check its validity for n=1. Assume that you have checked it out all the way for n = 1, 2, ..., k-1. Now use that assumption to show the validity for n = k.
That is assume that 1+4+...+(k-1)2 = (k-1)k(2k-1)/6 and then calculate 1+4+..+(k-1)2+k2 = (k-1)k(2k-1)/6 + k2 and verify that it is (k(k+1)(2k+1))/6.
661) The series
\sum_{r=1}^\infty \dfrac{1}{r^r}
has no closed form expression. However,
\int_0^1\dfrac{1}{x^x}\ \mathrm{d}x=\sum_{r=1}^\infty \dfrac{1}{r^r}
which can be evaluated to any degree of accuracy (incidentally the integral is not elementary). The value uptill 40 places of decimal is
1.291285997062663540407282590595600541499
23wat is d meaning of dis post bcomin pink??? does it mean it is d right ans?? who decides dis ???
23no!!!!!!!!!!! its correct ....actually i was asking whether the admin decide it or it is decided on the basis of votes ....
62The admins decide it..
a very few select student users also have the power.. (and i decide who has the rights to pink posts)
341It has a funny name: Sophomore's dream!
I think bhargav had asked a similar question some months back
Here it is: http://targetiit.com/iit-jee-forum/posts/16-02-2009-maths-a-great-combination-of-concepts-2463.html
