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Sum of squares of n terms

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Sum of squares of n terms

by sudhir3127 » Thu Jul 10, 2008 11:23 pm
hi all,

sorrie for starting a new thread for this question alone...
but was stuck with such a basic question... need ur help..

its 11^2+ 12^2+...........20^2

i know we have to use the formmula n(n+1)(2n+1)/6 but whats n here..

i read the explanation somewhere saying first we need to calculate sum of the square of 1 to 20 ( 1^2+ 2^2....20^2) and subratct from it (1^2+2^2....10^2)...

is there any other way 2 solive this ,,

thanks

Sudhir
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by sudhir3127 » Fri Jul 11, 2008 1:59 am
10 views and no answer... is the problem difficult or of such a low level?
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by gxliu » Fri Jul 11, 2008 5:31 am
post the question exactly along with the choices. its difficult trying to understand what you are asking.
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by sudhir3127 » Fri Jul 11, 2008 5:58 am
the question asks me to find the value of

11^2+12^2+13^2......+20^2

the answer choice are

a.385
b.2485
c.2870
d.3225

regards

Sudhir
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by gxliu » Fri Jul 11, 2008 6:31 am
Answer is B.
X=11
X^2+(X+1)^2+(X+2)^2...(X+9)^2
Comes out to 10X^2+90X+285
Sub 11 for X...works out to 2485
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Re: Sum of squares of n terms

by gabriel » Fri Jul 11, 2008 6:52 am
sudhir3127 wrote: i read the explanation somewhere saying first we need to calculate sum of the square of 1 to 20 ( 1^2+ 2^2....20^2) and subratct from it (1^2+2^2....10^2)...



Sudhir
This is exactly how you solve this problem by using the formula for summation of squares (i.e. the formula you mentioned in the post).

But I liked the method posted by gxliu better.
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by sudhir3127 » Fri Jul 11, 2008 7:04 am
thanks a lot ..can please explain how did u deduce to the eqn 10X^2 +90X+285... any formula?
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by gxliu » Fri Jul 11, 2008 7:07 am
look for the pattern as you work out the first 3 terms of the sequence.
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by sudhir3127 » Fri Jul 11, 2008 7:52 am
still confused.. can u please solve and help... i am not finding any series .. i mean i dont find it to be AP series.. as "D" is not constant..
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by lion147 » Fri Jul 11, 2008 8:14 am
gxliu wrote:Answer is B.
X=11
X^2+(X+1)^2+(X+2)^2...(X+9)^2
Comes out to 10X^2+90X+285
Sub 11 for X...works out to 2485
Thanks, took me a while to work that out, but I think that's faster.

It's probably helpful to know where the 285 comes from too.
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by ildude02 » Fri Jul 11, 2008 8:15 am
Can we use the formulae n(n+1)(2n+1)/6 at all ? I mean, this seems to be a series of squares with n = 10;

If we do use this in the formulae,

we get 10 * 11 * 21 / 6 = 385. Obviously that's worng. So I'm wondering when can we use this sum of sqaures formulae at all? Appreciate anyone's response. Can this only be used if the terms of squares starting from 1 and are consecutive? such as 1 ^2 + 2 ^2 + 3 ^ 2........ETC? Is it the same thing even with the sum of concesutive series , n(n+1)/2 ? Can we ONLY apply them if the series starts with "1" ?
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by tycoon_316 » Fri Jul 11, 2008 8:31 am
u can solve this using the formula n ( n+1) ( 2n+1)/6

1st caluclate for n=10 which would give u sum of squares from 1 to 10
2nd calculate for n=20 which would give you sum of squares from 1 to 20


1st= 385 for n =10
2nd = 2870 for n =20

subtract 1st from 2nd for the answer 2870-385=2485 which is choice b

hope u get it cheers
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by ildude02 » Fri Jul 11, 2008 10:26 am
thanks. So we need to always consider n from 1 and dependign on what the question series is, subtract it appropriately? I mean, if the question was for 20 ^2 .....30 ^2, then calculate from 1 to 20 square and subtract it from 1 to 30 square, right?

Is this the same concept for even the sum of concesutive integers? Always calculate based off 1?
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by gabriel » Fri Jul 11, 2008 3:13 pm
ildude02 wrote:thanks. So we need to always consider n from 1 and dependign on what the question series is, subtract it appropriately? I mean, if the question was for 20 ^2 .....30 ^2, then calculate from 1 to 20 square and subtract it from 1 to 30 square, right?

Is this the same concept for even the sum of concesutive integers? Always calculate based off 1?
Yes that is right. Remember all the summation formulas are meant for consecutive positive integers starting from 1. So if you have a situation where you have to find the sum of numbers from 31^2 to 40^2 .. then you first calculate the sum of 1^2 to 40^2 using the formula and then subtract from it the sum of 1^2 to 30^2. Hope this helps.

Regards
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