What a fascinating year we are in!
Given that the limit above can be expressed as , where and are coprime positive integers, find .
The answer is 2017.
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1 solution
x → 1 lim x 2 − 1 x 2 0 1 7 − 2 x + 1 <=> x → 1 lim x 2 − 1 x ( x 2 0 1 6 − 1 ) − ( x − 1 )
applying the formula a n − 1 = ( a − 1 ) ( a n − 1 + a n − 2 + . . . + a + 1 ) <=> x → 1 lim ( x − 1 ) ( x + 1 ) x ( x − 1 ) ( x 2 0 1 5 + x 2 0 1 4 + . . . + x + 1 ) − ( x − 1 ) <=> x → 1 lim ( x + 1 ) x ( x 2 0 1 5 + x 2 0 1 4 + . . . + x + 1 ) − 1 = 1 + 1 1 ( 1 2 0 1 5 + 1 2 0 1 4 + . . . + 1 + 1 ) − 1 So we can conclude that x → 1 lim x 2 − 1 x 2 0 1 7 − 2 x + 1 = 2 2 0 1 5