Question

I need to do a summation problem without a calculator. Can someone remind me the tricks on how to do this kind of problem?

Answer 1

The problem is \[\sum_{x = 1}^{53}j(j-1)\]

Answer 2

sorry, I meant j = 1 not x

Answer 3

so the number 53 means the numbers of terms.

Answer 4

j=1 means what term you start with.

Answer 5

So you can substitute j with 1. That would be your first term. Then you can then sub in j=2 and so on, all the way up to j=53.

Answer 6

But for now you should sub in j=1, j=2 and j=53

Answer 7

I'm supposed to be able to do this without a calculator. Which I know this one isn't too hard but aren't there formulas or shortcuts to get the answer?

Answer 8

Yes. But the first step is to do that, so you know your pattern.

Answer 9

Never skip steps. Always show all working, so markers know what you're doing.

Answer 10

add the firt and last

divide by 2

multiply result by the number of terms.

Answer 11

Well j=1 = 0, j=2 =2, j=53 = 2756

Answer 12

No...

Answer 13

Luis, that gives me 1431, the answer is supposed to be 49,608.

Answer 14

\[1(0)+2(1)+3(2)...+53(52)\]

Answer 15

See theere's two patterns.

Answer 16

there's*

Answer 17

hey, in SAT, GMAT, ect, yoy have to skip steps. I should know

Answer 18

Yeah well you got it wrong.

Answer 19

It's not even an A.P.

Answer 20

That's only half of the pattern.

Answer 21

\[j(j-1)\]
There's two patterns. Separate the J and the J-1.
(1+2+3+...+53)
(0+1+2+...+52)
We will multiply the sum of the two A.P's at the end to get what you're looking for.

Answer 22

I know that. I'm asking if there are formulas to help. Like if it was just j and not j(j-1). I could've used (n(n+1))/2

Answer 23

Wait sorry about that.

Answer 24

It's
\[j^2-j\]

Answer 25

OHHH! I can't believe I didn't catch that!!!! THANKS BRO!

Answer 26

So the pattern would be like this.
\[1-1+4-2+9-3...+2809-53\]

Answer 27

Alternating

Answer 28

So G.P and an A.P

Answer 29

or I could split the summation into two summations. Then do the formulas for summations of j^2 and j

Answer 30

Do them separately. Add them together. I'm sure that the A.P will have a negative sum.

Answer 31

That's what I said.

Answer 32

Remember to use the G.P summation with the G.P and the A.P summation with the A.P.

Answer 33

Most people use just one of the summation equations for both.