The sequence {ak}^112(upper) & k=1 (lower) satisfies a1=1 and a_n = (1337+n)/(a_n-1) , for all positive integers n. Let S=⌊ a_10*a_13+a_11*a_14+a_12*a_15+⋯+a_109*a_… ⌋
Find the remainder when S is divided by 1000
look on the picture
http://imageshack.us/photo/my-images/211/seqz.jpg/
Find the remainder when S is divided by 1000
look on the picture
http://imageshack.us/photo/my-images/211/seqz.jpg/
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S = 139849
remainder S/1000 is 849
S = floor(a_10*a_13+a_11*a_14+a_12*a_15+⋯+a_…
a_10*a_13 ≈ 1348.99
a_109*a_112 ≈ 1447.99
1349 + 1350 + ... + 1448 = 139850
a_10*a_13+a_11*a_14+a_12*a_15+⋯+a_109*… ≈ 139849.93
the floor of this number is 139849 therefore the remainder when S is divided by 1000
is 849
remainder S/1000 is 849
S = floor(a_10*a_13+a_11*a_14+a_12*a_15+⋯+a_…
a_10*a_13 ≈ 1348.99
a_109*a_112 ≈ 1447.99
1349 + 1350 + ... + 1448 = 139850
a_10*a_13+a_11*a_14+a_12*a_15+⋯+a_109*… ≈ 139849.93
the floor of this number is 139849 therefore the remainder when S is divided by 1000
is 849