QED - Rose
... For convenience, lets assume that ak 0 , so that n P(k ) . If the first term was zero, all that would be required would be to consider a smaller number of terms and re-labeling subscripts accordingly. We can break the inductive step up into three cases. Case I: i k st ai p(i 1) 1 (This ...
... For convenience, lets assume that ak 0 , so that n P(k ) . If the first term was zero, all that would be required would be to consider a smaller number of terms and re-labeling subscripts accordingly. We can break the inductive step up into three cases. Case I: i k st ai p(i 1) 1 (This ...
Full text
... If there are no more l f s to be changed at the end of a loop, the Markov algorithm stops at rule 12, indicating that the original string of l f s was a Fibonacci number. If, however, the string was not a Fibonacci number, the Markov algorithm jumps out of the loop in midstream of changing l's to a ...
... If there are no more l f s to be changed at the end of a loop, the Markov algorithm stops at rule 12, indicating that the original string of l f s was a Fibonacci number. If, however, the string was not a Fibonacci number, the Markov algorithm jumps out of the loop in midstream of changing l's to a ...
Quantitative Ability – POINTS TO REMEMBER If an equation (i.e. f(x
... 17. a4 + b4 + c4 + d4 >= 4abcd (Equality arises when a=b=c=d=1) 18. (n!)2 > nn 19. If a + b + c + d=constant, then the product a^p * b^q * c^r * d^s will be maximum if a/p = b/q = c/r = d/s 20. If n is even, n(n+1)(n+2) is divisible by 24 21. x^n -a^n = (x-a)(x^(n-1) + x^(n-2) + .......+ a^(n-1) ) . ...
... 17. a4 + b4 + c4 + d4 >= 4abcd (Equality arises when a=b=c=d=1) 18. (n!)2 > nn 19. If a + b + c + d=constant, then the product a^p * b^q * c^r * d^s will be maximum if a/p = b/q = c/r = d/s 20. If n is even, n(n+1)(n+2) is divisible by 24 21. x^n -a^n = (x-a)(x^(n-1) + x^(n-2) + .......+ a^(n-1) ) . ...