1. ## 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!

2. 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,

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

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

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

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

Unfortunately the last example isn't unambiguous:

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

2. $\displaystyle f(x)=(2x^7 - x^2)\cdot \left( \frac{x - 1}{x + 1}\right) ~\longrightarrow~ f'(x) = \frac{2x\left(7x^7+2x^6-7x^5-x^2-x+1\right)} {(x+1)^2}$

3. Hello, Mr.O!

Evaluate: .$\displaystyle \sum^{10}_{i=1}i(i^2+1)$
If this is the first lecture on Summations, you may not know any formulas yet.
. . You are probably expected to find the sum directly . . .

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

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

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