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(4*12+4*22+4*32+…+4*102)/(3*1+3*2+…+3*10)=?

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(4*12+4*22+4*32+…+4*102)/(3*1+3*2+…+3*10)=?

by Max@Math Revolution » Mon May 02, 2016 10:36 pm
(4*12+4*22+4*32+...+4*102)/(3*1+3*2+...+3*10)=?

A. 71/9
B. 79/9
C. 28/3
D. 93/9
E. 125/9

* A solution will be posted in two days.
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by Max@Math Revolution » Thu May 05, 2016 9:37 pm
We know that 1^2+2^2+....+n^2=n(n+1)(2n+1)/6 and 1+2+....+n=n(n+1)/2. Then, from the question, the numerator becomes 4(1^2+2^2+..10^2)=4(10)(11)(21)/6 and the denominator becomes 3(1+2+....+10)=3(10)(11)/2. Hence, (4*1^2+4*2^2+4*3^2+...+4*10^2)/(3*1+3*2+...+3*10) =[4(10)(11)(21)/6]/[3(10)(11)/2]=28/3. Hence, the correct answer is C.
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