Bunuel wrote:
What is the value of 27^2 + 28^2 + 29^2 + 30^2 + 31^2 + 32^2 + 33^2 ?
A. 6298
B. 6308
C. 6318
D. 6328
E. 6338
hi,
the choices being very close, any guess work may not be fruitful...
1)
a straight formula for this would be \(\frac{n(n+1)(2n+1)}{6}\).....
we have to find \(1^2+2^2......+33^2-(1^2+2^2......+27^2)\)
so n in two case is 33 and 26...
\(\frac{33*34*67}{6}-(\frac{26*27*53}{6})\).... thereafter we require calculations which will give us 6328 as the ans...
2)the method would seem to be lengthy as it is being explained but is very short actually second would be to take advantage of proximity to 30 and a set of two numbers giving us 60 as sum..
\((27^2 + 33^2) + (29^2 + 31^2) + (32^2 + 28^2)+ 30^2\)...
\((60^2-2*27*33) +(60^2-2*29*31)+ (60^2-2*28*32)+ 30^2\)..
\(3*60^2+30^2-2(27*33+29*31+28*32)\)..
\(3*60^2+30^2-2((30-3)*(30+3)+(30-1)*(30+1)+(30-2)*(30+2))\)..
\(3*60^2+30^2-2((30^2-3^2)+(30^2-1^2)+(30^2-2^2))\)..
\(3*60^2+30^2-6(30^2)+
2(3^2+1^2+2^2)\)
we are interested only in terms where square of multiple of 10 is not involved..2(9+1+4)=28 so the answer should be something*100+28....this gives us the answer as 6328
ans D
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