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# IB maths question about sequence and sigma .

find the sum of (Ui * Vi) until i  = 10 , which means i = 1,2,3,4,....10.

Ui = -3+4i      Vi = 12-3i

I know the answer is -1845 but i don't have a clue where it came from.   Please suggest...

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You're trying to find the sum,

$\sum_{i=0}^{10} U_i V_i = \sum_{i=0}^{10} (-3+4i)(12-3i)$

One way that comes to mind is to note the linearity property of summation, namely that,

$\sum_{i=1}^{10} (ai^2 + bi + c) = a\left(\sum_{i=1}^{10} i^2\right) + b \left(\sum_{i=1}^{10} i\right) + c \left(\sum_{i=1}^{10} 1\right)$

Where $a,b,c$ are constants. Calculating the sum of the sequence $i^2$ isn't exactly easy unless you know a specific identity to do it (which isn't in the IB syllabus), but that does simplify the calculation a fair bit (though it is still fairly messy, none of the multiplications are particularly nice).

I'm not too sure how else you can easily sum this without a calculator, where did you get this question from?

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You're trying to find the sum,

$\sum_{i=0}^{10} U_i V_i = \sum_{i=0}^{10} (-3+4i)(12-3i)$

One way that comes to mind is to note the linearity property of summation, namely that,

$\sum_{i=1}^{10} (ai^2 + bi + c) = a\left(\sum_{i=1}^{10} i^2\right) + b \left(\sum_{i=1}^{10} i\right) + c \left(\sum_{i=1}^{10} 1\right)$

Where $a,b,c$ are constants. Calculating the sum of the sequence $i^2$ isn't exactly easy unless you know a specific identity to do it (which isn't in the IB syllabus), but that does simplify the calculation a fair bit (though it is still fairly messy, none of the multiplications are particularly nice).

I'm not too sure how else you can easily sum this without a calculator, where did you get this question from?

This is exactly what I thought as well. If this question was in indeed IB's, then it would probably be a calculator question. On the other hand though, the sum of the squares can be simplified by using this identity (from http://www.trans4mind.com/personal_development/mathematics/series/sumNaturalSquares.htm):

$\sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}$; in other words $\sum_{i=1}^{10} i^2 = \frac{10(11)(21)}{6} = 385$

With this identity, the calculations can be easily done by hand; but of course this is not in the syllabus, so don't even bother. Use your Ti-84

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I have attached my solution.

Regards

Aniruddh

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I have attached my solution.

Regards

Aniruddh

You're trying to find the sum,

$\sum_{i=0}^{10} U_i V_i = \sum_{i=0}^{10} (-3+4i)(12-3i)$

One way that comes to mind is to note the linearity property of summation, namely that,

$\sum_{i=1}^{10} (ai^2 + bi + c) = a\left(\sum_{i=1}^{10} i^2\right) + b \left(\sum_{i=1}^{10} i\right) + c \left(\sum_{i=1}^{10} 1\right)$

Where $a,b,c$ are constants. Calculating the sum of the sequence $i^2$ isn't exactly easy unless you know a specific identity to do it (which isn't in the IB syllabus), but that does simplify the calculation a fair bit (though it is still fairly messy, none of the multiplications are particularly nice).

I'm not too sure how else you can easily sum this without a calculator, where did you get this question from?

This is exactly what I thought as well. If this question was in indeed IB's, then it would probably be a calculator question. On the other hand though, the sum of the squares can be simplified by using this identity (from http://www.trans4mind.com/personal_development/mathematics/series/sumNaturalSquares.htm):

$\sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}$; in other words $\sum_{i=1}^{10} i^2 = \frac{10(11)(21)}{6} = 385$

With this identity, the calculations can be easily done by hand; but of course this is not in the syllabus, so don't even bother. Use your Ti-84

You're trying to find the sum,

$\sum_{i=0}^{10} U_i V_i = \sum_{i=0}^{10} (-3+4i)(12-3i)$

One way that comes to mind is to note the linearity property of summation, namely that,

$\sum_{i=1}^{10} (ai^2 + bi + c) = a\left(\sum_{i=1}^{10} i^2\right) + b \left(\sum_{i=1}^{10} i\right) + c \left(\sum_{i=1}^{10} 1\right)$

Where $a,b,c$ are constants. Calculating the sum of the sequence $i^2$ isn't exactly easy unless you know a specific identity to do it (which isn't in the IB syllabus), but that does simplify the calculation a fair bit (though it is still fairly messy, none of the multiplications are particularly nice).

I'm not too sure how else you can easily sum this without a calculator, where did you get this question from?

OMG I am so glad that it is not in the IB syllabus . I locked myself in my room for 2 hours , trying to solve that stupid question. Now i am  happy.

By the way  i got it from my IBID maths HL book .... weird weird

But thank  you sooooo much.

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No one in my class got it either but then our math teacher taught us how to do it.

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