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
Sum of squares of n terms
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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
11^2+12^2+13^2......+20^2
the answer choice are
a.385
b.2485
c.2870
d.3225
regards
Sudhir
- gabriel
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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).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
But I liked the method posted by gxliu better.
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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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Thanks, took me a while to work that out, but I think that's faster.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
It's probably helpful to know where the 285 comes from too.
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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" ?
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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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
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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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?
Is this the same concept for even the sum of concesutive integers? Always calculate based off 1?
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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.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?
Regards