How do I tackle these problems?

Question

usukidoll · Answer

attachment

dape · Answer

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 · Answer

looks familiar like a sequence. ....

dape · Answer

Disregard my lack of grammatical correctness, i'm typing on an ipad and it's 5am.

usukidoll · Answer

attachment

usukidoll · Answer

am I on the right track?

missmob · Answer

i got nothing

usukidoll · Answer

see attachment..it's too long

usukidoll · Answer

click on sixone.png and then the original problem.

missmob · Answer

i cant understand wat u wrote there

usukidoll · Answer

|dw:1377592856047:dw|

missmob · Answer

i think it will be better to ask @mathslover

usukidoll · Answer

k k

usukidoll · Answer

@mathslover am I on the right track with this?

mathslover · Answer

Well, sorry, I don't think so...

mathslover · Answer

$$\sum_{i = 1}^{3}a = 1 + 2 + 3 = $$

mathslover · Answer

Similarly solve for : $$\sum_{i=3}^{20} i = ?$$

mathslover · Answer

or 3 + 4 + ... + 20 = (1+2) + 3 + ... + 20 - (1+2)

mathslover · Answer

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 · Answer

so I just use n(n+1)/2 and n=20?

usukidoll · Answer

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 · Answer

|dw:1377593886467:dw|
2870-3=2867