SOLUTION: Find the sum of 10 terms of the series log3+log6+log12+... given that log3 = 0.47741 and log2+0.3010

Question 1146289: Find the sum of 10 terms of the series log3+log6+log12+... given that log3 = 0.47741 and log2+0.3010
Answer by greenestamps(13203) (Show Source): You can put this solution on YOUR website!

log3+log6+log12+... (to 10 terms) =
log3+log(3*2)+log(3*2^2)+...+log(3*2^9) =
log3+(log3+log2)+(log3+log(2^2))+...+(log3+log(2^9)) =
log3+(log3+log2)+(log3+2log2)+...+(log3+9log2) =
10log3+(1+2+...+9)log2 =
10log3+45log2 =
10(0.47741)+45(0.3010) =
4.7741+13.545 =
18.3191
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