UsukiDoll
  • UsukiDoll
How do I tackle these problems?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
UsukiDoll
  • UsukiDoll
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dape
  • dape
The sum of i from 1 to n is n(n+1)/2, by adding and subtracting the same numbers you can getthis sum to start from 1 and use this formula. Similarly the \(i^2\) sum can be written as be written as n(n+1)(2n+1)/6. You can prove these formulaa by induction if you don't believe me.
UsukiDoll
  • UsukiDoll
looks familiar like a sequence. ....

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dape
  • dape
Disregard my lack of grammatical correctness, i'm typing on an ipad and it's 5am.
UsukiDoll
  • UsukiDoll
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UsukiDoll
  • UsukiDoll
am I on the right track?
missMob
  • missMob
i got nothing
UsukiDoll
  • UsukiDoll
see attachment..it's too long
UsukiDoll
  • UsukiDoll
click on sixone.png and then the original problem.
missMob
  • missMob
i cant understand wat u wrote there
UsukiDoll
  • UsukiDoll
|dw:1377592856047:dw|
missMob
  • missMob
i think it will be better to ask @mathslover
UsukiDoll
  • UsukiDoll
k k
UsukiDoll
  • UsukiDoll
@mathslover am I on the right track with this?
mathslover
  • mathslover
Well, sorry, I don't think so...
mathslover
  • mathslover
\[\sum_{i = 1}^{3}a = 1 + 2 + 3 = \]
mathslover
  • mathslover
Similarly solve for : \[\sum_{i=3}^{20} i = ?\]
mathslover
  • mathslover
or 3 + 4 + ... + 20 = (1+2) + 3 + ... + 20 - (1+2)
mathslover
  • mathslover
Now solve for : (1 + 2 + ... + 20) - (1+2) 1 + 2 + ... + 20 = n(n+1) /2 = 20(21)/2= 210 so you get : 210 - 3 = 207
UsukiDoll
  • UsukiDoll
so I just use n(n+1)/2 and n=20?
UsukiDoll
  • UsukiDoll
what about for I^2 would it be n(n+1)(2n+1)/6 and n = 20 and then after I get that I subtract 3?
UsukiDoll
  • UsukiDoll
|dw:1377593886467:dw| 2870-3=2867
UsukiDoll
  • UsukiDoll
@mathslover
mathslover
  • mathslover
Sorry was away for sometime, yep u r right
UsukiDoll
  • UsukiDoll
YES!!! *HHUGS*!
mathslover
  • mathslover
oops wait, 2870 - (1^2 + 2^2 + 3^2)
UsukiDoll
  • UsukiDoll
oh crud just realized it too...
UsukiDoll
  • UsukiDoll
2856
mathslover
  • mathslover
3^2 + ... + 20^2 = 1^2 + 2^2 + 3^2 + ... +20^2 - (1^2 + 2^2) which is ?
mathslover
  • mathslover
we made a mistake again : it should be : \((1^2 + 2^2) + (3^2 + ... + 20^2) - ( 1^2 + 2^2) = 2870 - 5 = 2865 \)

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