Question 264758: 3+6+9+...+996+999=
1+3+3^2+3^3+...+3^30=
1+(-1)+(-1)^2+...+(-1)^1000=
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! 3+6+9+...+996+999=
An arithmetic sequence with a(1) = 3 ; d = 3
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Find n where a(n) = 999
999 = 3 + (n-1)3
996 = (n-1)3
n-1 = 332
n = 333
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Find the sum:
S(333) = (333/2)*(3+999)= 333*1002/2 = 333*501 = 166,833
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1+3+3^2+3^3+...+3^30=
A geometric sequence with n = 31 and r = 3
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S(31) = (3^31 - 1)/(3-1) = 3.088..x10^14
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1+(-1)+(-1)^2+...+(-1)^1000=
Number of terms = 2000
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# of pairs = 1000
Sum of each pair = 0
Total sum = 0
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Cheers,
Stan H.
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