n → ∞ lim ( ( n + 1 ) 2 n + ( n + 2 ) 2 n + ( n + 3 ) 2 n + . . . + 2 5 n 1 )
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Using the limit definition of an integral:
S = n → ∞ lim ( ( n + 1 ) 2 n + ( n + 2 ) 2 n + ( n + 3 ) 2 n + . . . + 2 5 n 1 ) =
n → ∞ lim i = 1 ∑ 4 n ( n + i ) 2 n = n → ∞ lim n 1 i = 1 ∑ 4 n ( 1 + n i ) 2 1 = ∫ 0 4 ( 1 + x ) 2 1 d x = 5 4
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Note that the limit can be re-written as From a Riemann Sum, we know that So,