summations

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December 17th 2007, 11:29 PM
Mr.Obson

summations

my teacher was in a rush at the end of the class and he literally only spent 5 minutes on summations, so i'm kinda confused here.
Evaluate:
10
Σi(i^2+1)
i=1

Evaluate:
10
Σk(k- 2)(k+2)
k=1
oh and also if you can, i think i got it right, but i always make some dumb mistakes on these long ones

find the derivative y = (2x^7 - x^2)(x - 1/x + 1)
just the answer would be enough for this last one, if it doesn't match then i'll check my math again.
Love ya guys!
December 18th 2007, 01:57 AM
earboth

Quote:

Originally Posted by Mr.Obson

my teacher was in a rush at the end of the class and he literally only spent 5 minutes on summations, so i'm kinda confused here.
Evaluate:
10
Σi(i^2+1)
i=1

Evaluate:
10
Σk(k- 2)(k+2)
k=1
oh and also if you can, i think i got it right, but i always make some dumb mistakes on these long ones

find the derivative y = (2x^7 - x^2)(x - 1/x + 1)
just the answer would be enough for this last one, if it doesn't match then i'll check my math again.
Love ya guys!

Hello,

$\sum^{10}_{i=1}\left(i (i^2+1)\right) = \sum^{10}_{i=1}\left(i^3+i \right) = \sum^{10}_{i=1}\left( i^3\right) + \sum^{10}_{i=1}\left(i \right)$ . Use the formulae for these sums:

$\sum^{10}_{i=1}\left(i (i^2+1)\right) = \frac{10^2(10 + 1)^2}{4} + \frac12 \cdot 10(10+1) = 3080$

The second example can be done in just the same way:

$\sum^{10}_{k=1} \left(k (k+2)(k-2) \right) = \sum^{10}_{k=1}\left(k^3 - 4k\right)$ . Now continue.

Unfortunately the last example isn't unambiguous:

1. $f(x)= (2x^7 - x^2) \left(x - \frac1x + 1 \right)~\longrightarrow~ f'(x)=16x^7 + 14x^6 -12x^5 -3x^2 - 2x + 1,~x \neq 0$

2. $f(x)=(2x^7 - x^2)\cdot \left( \frac{x - 1}{x + 1}\right) ~\longrightarrow~ f'(x) = \frac{2x\left(7x^7+2x^6-7x^5-x^2-x+1\right)} {(x+1)^2}$
December 18th 2007, 06:50 AM
Soroban

Hello, Mr.O!

Quote:

Evaluate: . $\sum^{10}_{i=1}i(i^2+1)$

If this is the first lecture on Summations, you may not know any formulas yet.
. . You are probably expected to find the sum directly . . .

$\sum^{10}_{i=1}i(i^2+1) \;=\;1(1^2+1)\;+\;2(2^2+1)\;+\;3(3^2+1)\;+\;4(4^2+ 1)$ $+\cdots+\;9(9^2+1)\;+\;10(10^2+1)$

. . . . . . . . $= \;1\!\cdot\!2\;+\;2\!\cdot\!5\;+\;3\!\cdot\!10\;+\ ;4\!\cdot\!17\;+\;\cdots\;+\;9\!\cdot\!82\;+\;10\! \cdot\!101$

. . . . . . . . $= \;3080$