MathRevolution wrote:
(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.
Hi,
two points..
1) it is not a MIXTURE problem..
2) (4*12+4*22+4*32+…+4*102) is not likely to give you 11 on numerator..
You are missing out on some signs.. as 12 may be 1^2.. pl recheck
The Q is supposed to be --
(4*12+4*22+4*32+…+4*102)/(3*1+3*2+…+3*10)= \(\frac{(4*1^2+4*2^2+4*3^2+…+4*10^2)}{(3*1+3*2+…+3*10)}\)
\(\frac{4(1^2+2^2+...+10^2)}{3(1+2+3...+10)} = \frac{4*10*(10+1)(2*10+1)}{6}/ \frac{3*10*11}{2}\)
=> \(\frac{4*10*11*21*2}{3*10*11*6} = \frac{28}{3}\)
C..
Ofcourse you require to know that \(1^2+2^2+...n^2 = \frac{n(n+1)(2n+1)}{6}\)
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