What is the value of 27^2+28^2+29^2+30^2+31^2+32^2+33^2

I am stuck to 6328!
This is what I did, all the nos are close to 30, and squared, so 30^2 plus there are 7 numbers, so (30)^2 x 7 =6300,
now squaring the last digits of each no individually comes up to (9+4+1+0+1+4+9) = 28.

when you add the 2 up it gives you a (6300+28) =6328, Answer is D IMO.

Last 2 digit of 27^2 is=29
28^2=84
29^2=41
30^2=00
31^2=61
32^2=24
33^2=89

Sum them up to get 328. Answer D.

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

i found this trick useful in exam so just want to share it..
how to get last 2-digits in multiplication of 2 two digit numbers.

took a different route :

27^2 + 28^2 + 29^2 + 30^2 + 31^2 + 32^2 + 33^2

arranged the series in the form pairs of 60 .....(27+33) , (28+32) , (29+31) and using formula \(a^2+b^2 = (a+b)^2 - 2ab\)

\((27+33)^2 - 2 (27 . 33) + (28+32)^2 - 2 (28 . 32) + (29+31)^2 - 2 (29 . 31) + 30^2\)

\((27+33)^2 + (28+32)^2 + (29+31)^2 - 2 (27 . 33 + 28 . 32 + 29 . 31 ) + 30^2\)

\(3(60)^2 - 2 (27 . 33 + 28 . 32 + 29 . 31 ) + 30^2\)

=3 (3600) - 2 ( 2686) + 900

=10800 - 5372 + + 900 = 6328

Alternate way , I like the way Jackal did it

\((30-3)^2 + (30-2)^2 + (30-1)^2 + (30)^2 + (30+3)^2 + (30+2)^2 + (30+1)^2\)

\(7(30^2) + 2(3^2) + 2(2^2) + 2(1^2) + (2.30.3 + 2.30.2 + 2.30.1 - 2.30.3 - 2.30.2 - 2.30.1)\)

=7*30^2 + 28 = 6328

Ans D

27^2 + 28^2 + 29^2 + 30^2 + 31^2 + 32^2 + 33^2

We can solve by taking pairs:
29^2 = (30-1)^2 = 30^2 - (2)(30)(1) + 1^2
31^2 = (30+1)^2 = 30^2 + (2)(30)(1) + 1^2
The sum is 30^2+30^2+2

28^2 = (30-2)^2 = 30^2 - (2)(30)(2) + 2^2
32^2 = (30+2)^2 = 30^2 +(2)(30)(2) + 2^2
The sum is 30^2+30^2+8

27^2 = (30-3)^2 = 30^2 - (2)(30)(3) + 3^2
33^2 = (30+3)^2 = 30^2 + (2)(30)(3) + 3^2
The sum is 30^2+30^2+18

So we got 7 of the 30^2 and 2+8+18
900(7)+28=6328

Answer D

I would use the last digit analysis

7^2=49 - 9 is the last digit
8^2=64 - 4 is the last digit
9^2=81 - 1 is the last digit
0^2=0 0 is the last digit
1^2 =1- 1
2^2=4 -4
3^2=9- 9

Take the units digit and then just add 9+4+1+0+1+4+9 =28 --> Check the answers only 6328 matches !!

Easy and quick !!

how you calculated the last 2 digits , please explain the method used.

This makes no sense. Just right in this case.

In problems that are computation intensive, one clue could be symmetry. Here we see on the left of 30^2,
(30-1)^2, (30-2)^2 and (30-3)^2 and on the right (30+1)^2. (30+2)^2 and (30 +3)^2

We can easily see the form (a-b)^2 and (a+b)^2 and that -2ab and +2ab terms cancel out.
Sum of a^2 terms is 6300 and b^2 terms is 28. Total is 6328.

When I see this, I remember that the difference between one square and the next has a progression of 2.

30^2 = 900
31^2 = 961
Difference is 61. The next difference will be 63, i.e. 32^2 - 31^2 = 63.

The sum gets:

900*7
+ 61*3 + 63*2 + 65
- 59*3 - 57*2 - 55

6300 + 2*3 + 6*2 + 10 = 6328

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