Question

Sigma Notation Question Help

Answer 1

a) Given that \[\sum_{r=1}^{n}r= \frac{ n(n+1) }{ 2 }, express \sum_{r=1}^{n}\left[ r ^{2}-(r-1)^{2} \right] interms of n\]
b) Hence, or otherwise, evaluate \[\sum_{r=51}^{100}\left[ r ^{2}-(r-1)^{2} \right]\]

Answer 2

can you simplify
r^2 - (r-1)^2 ??

Answer 3

If I simplify I will get 2r-1

Answer 4

correct!

Answer 5

so, \(\sum (2r-1) = \sum 2r - \sum 1 = 2 \sum r - \sum 1 \)

Answer 6

you know what \(\sum \limits_{r=1}^n1 =... ?\)

Answer 7

Um..1 x n?

Answer 8

yes!

Answer 9

so, can you solve further now ?

Answer 10

Nope, the solving part is where I'm getting problems.

Answer 11

How will I write that 1 x n bit?

Answer 12

\(2\sum r - \sum 1 = 2 \dfrac{n(n+1)}{2}-n = ...?\)

Answer 13

1 x n = n

Answer 14

\[=n ^{2}\] ??

Answer 15

correct!

Answer 16

So is that the answer for part a?

Answer 17

absolutely :)

Answer 18

What about part b? Can you help me with?

Answer 19

sure

Answer 20

the part b asks for sum of 'r' values from 51 to 100
what we can do instead is , find sum of 'r' values from 1 to 100 and SUBTRACT it from 'r' values from 1 to 50
that should give us sum of 'r' values from 51 to 100
makes sense ?

Answer 21

\(\sum \limits_{r= 51}^{100} =\sum \limits_{r= 1}^{100}-\sum \limits_{r=1}^{50} \)

Answer 22

Yes, but I want to know how I can expand out to get the answer

Answer 23

once you understand this, its just using the previous result :)
in which we got
\(\sum \limits_{r=1}^{n}... = n^2\)

so,
\(\sum \limits_{r= 1}^{100}... = 100^2\)
and
\(\sum \limits_{r=1}^{50}...=50^2\)

Answer 24

got this part ?

Answer 25

Oh yeah

Answer 26

\(\sum \limits_{r= 51}^{100}... =\sum \limits_{r= 1}^{100}...-\sum \limits_{r=1}^{50}... = 100^2-50^2=....?\)
calculate it, and thats it! :)

Answer 27

7,500..but isn't it 51 and not 50?

Answer 28

yes, its 51
see, if it was from, 4 to 7
i would have taken
1 to 7 - 1 to 3
right ?
4,5,6,7 = 1,2,3,4,5,6,7 - 1,2,3

Answer 29

and not 1 to 7 - 1 to 4
because that will give me 5,6,7

Answer 30

makes sense ?

Answer 31

Yup, I understand now. But can you show me how I will expand out the sigma for part b?

Answer 32

i did
\(\sum \limits_{r= 51}^{100}... =\sum \limits_{r= 1}^{100}...-\sum \limits_{r=1}^{50}... = 100^2-50^2=7500\)
just write
\([r^2 -(r-1)^2]\)
instead of ...

Answer 33

Thank you! :)

Answer 34

welcome ^_^

Answer 35

I typed in part b in a solver and the answer they gave me was 295,425?

Answer 36

7500 is actually correct
http://www.wolframalpha.com/input/?i=sum+from+r+%3D+51+to+100+%28r%5E2+-+%28r-1%29%5E2%29
maybe you have done some error in giving the input to solver ?

Answer 37

sorry for the late reply, you know that we are not getting the notification, so i didn't know you replied to this