An algebra problem by Subrata Dutta
M=8^2+9^2+10^2+.....14^2+22^2+23^2+24^2+.....+28^2+.....................+120^2+121^2+...........+126^2 it can be written as , M/7^2=a+b/7 , then , a+b=?
The answer is 7009.
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1 solution
S=-[1^2+2^2+.....+r^2]+[(r+1)^2+(r+2)^2+......+(2r)^2]-........[(nr+1)^2+(nr+2)^2+........(nr+r)^2] =4r^3[i+1/4r+1/2] where , i=0,1,2,3,...... and 'n' must be odd and also , i=(n-1)/2 so, here, putting, r=7, n=17, we get, i=8, therefore S=-[1^2+2^2+......7^2]-[8^2+9^2+......+14^2]+..............[120^2+121^2+......+126^2] =4 7^3[8+1/(4 7)+1/2]=11711 again T=1^2+2^2+3^2+..........................+126^2=674751 so, M=(S+T)/2=343231, M/7^2=7004+5/7, a=7004, b=5, so the ans is, a+b=7004+5=7009