Let k and n be positive integers and put Sk=Σnk (n varies from 1 to n.
show that
m+1C1S1+m+1C2S2...............+m+1CmSm=(n+1)m+1-(n+1)
341(1+n)^{m+1} - n^{m+1} = 1 + \binom{m+1}{1} n + \binom{m+1}{2}n^2+....+\binom{m+1}{m} n^m
(n)^{m+1} - (n-1)^{m+1} = 1 + \binom{m+1}{1} (n-1) ++....+\binom{m+1}{m} (n-1)^m
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(2)^{m+1} - (1)^{m+1} = 1 + \binom{m+1}{1} (1) ++....+\binom{m+1}{m} (1)^m
Adding the above equations we get
(n+1)^{m+1} - (1) = n + \binom{m+1}{1} S_1 +\binom{m+1}{2}S_2 +....+\binom{m+1}{m} S_m
and hence we have(n+1)^{m+1} - (n+1) = \binom{m+1}{1} S_1 +\binom{m+1}{2}S_2 +....+\binom{m+1}{m} S_m
62phew.. only prophet sir can pull off this solution :)
seriously i am a big big fan of yours :)
24any other method exists ???
becoz i dont think this method will ever strike me in exam hall ....
3well no other method.And this wont strike me in the hall.If there is ne new method i will bookmark it for my revision.
24thats why I was aksing any other method ??
or if sir can explain us how did the first step strike his mind on seeing teh ques..then we can be benifitted...
341Before you guys shower more undeserved praises on me, let me tell you that this arises in the context of finding 1k+2k+...+nk by using our knowledge of the expressions for the sum for lower powers. You can look up Hall & Knight for this. I recognised this one bcos only this summer i was teaching my niece this stuff :D
24sir are u referring to article number 405 of hall and knight here ???