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# Sums of sequences, using sigma notation watch

1. If you need to find the sum of: 4 above the sigma sign, r=2 below the sigma sign, and r(r+1)?

I got 30, but the answer is 38? And i keep getting different values for d, the common difference??
2. 2(2+1) + 3(3+1) + 4(4+1) = 38
3. (Original post by *girlie*)
If you need to find the sum of: 4 above the sigma sign, r=2 below the sigma sign, and r(r+1)?

I got 30, but the answer is 38? And i keep getting different values for d, the common difference??
Sum (r = 2 to 4) r(r + 1) = Sum (r = 2 to 4) r^2 + r = Sum (r = 2 to 4) r^2 + Sum (r = 2 to 4) r = {Sum (r = 1 to 4) r^2 - 1^2} + {Sum (r = 1 to 4) r - 1} = (4/6)(5)(9) - 1 + (4/2)(5) - 1
= 30 - 1 + 10 - 1
= 40 - 2
= 38
4. Ohhh, thanks. But don't you just use the formula??

Erm for: 10 above sigma, r=5 below sigma, (3r+2), i keep getting 305 instead of 147?! Isn't d=3, a=17 and n=10?! What am i doing wrong??
5. you just plug the values 5, 6, 7, 8 ,9 and 10in (3r+2) and add them up.
6. Ohhh, i didn't realise that you had to do them separately...ok, this makes sense now! Thank you!
7. (Original post by *girlie*)
Ohhh, thanks. But don't you just use the formula??

Erm for: 10 above sigma, r=5 below sigma, (3r+2), i keep getting 305 instead of 147?! Isn't d=3, a=17 and n=10?! What am i doing wrong??
Sum (r = 5 to 10) 3r + 2 = 3 * Sum (r = 5 to 10) r + 2 * Sum (r = 5 to 10) 1 = 3{Sum r = 1 to 10) r - Sum (r = 1 to 4)} + 2[10 - 4] = 3{5(11) - 2(5)] + 12 = (45 * 3) + 12 = 147
8. (Original post by *girlie*)
If you need to find the sum of: 4 above the sigma sign, r=2 below the sigma sign, and r(r+1)?

I got 30, but the answer is 38? And i keep getting different values for d, the common difference??
Don't worry, the terms in the series do NOT form a LINEAR sequence,
they are quadratic, since there is an r^2.
So there wont be a common difference.
And you cant apply the usual sum of the series formula:
Sn=(n/2)[2a+(n-1)d], becuause it is NOT linear.

So adding the terms manually gives
2*3+3*4+4*5=6+12+20=38.

There is another way, but it is silly to use this in this question,
since the limits in the sum are small.
The formula SUM[r=1 to n]{r}=(1/2)*n(n+1)
and the formula SUM[r=1 to n]{r^2}=(1/6)*n(n+1)(2n+1)

So what you have is SUM[r=2 to 4]{r^2+r}=
SUM[r=1 to 4]{r^2+r}-SUM[r=1 to 1]{r^2+r}
=SUM[r=1 to 4]{r^2}+SUM[r=1 to 4]{r}-SUM[r=1 to 1]{r^2}-SUM[r=1 to 1]{r}
=(1/6)*4*5*9+(1/2)*4*5-(1/6)*1*2*3-(1/2)*1*2
=30+10-1-1=38.
9. (Original post by Nima)
Sum (r = 5 to 10) 3r + 2 = 3 * Sum (r = 5 to 10) r + 2 * Sum (r = 5 to 10) 1 = 3{Sum r = 1 to 10) r - Sum (r = 1 to 4)} + 2[10 - 4] = 3{5(11) - 2(5)] + 12 = (45 * 3) + 12 = 147
That way is well more complicated.
10. Wait a second...what do you do if the number above sigma is something like 40?! You dont just add them all up, do you?!
11. (Original post by *girlie*)
Wait a second...what do you do if the number above sigma is something like 40?! You dont just add them all up, do you?!
Hi, read my message above plz, just use the standard formulas
and remember SUM[r=1 to n]{A}=nA
12. (Original post by *girlie*)
Ohhh, thanks. But don't you just use the formula??

Erm for: 10 above sigma, r=5 below sigma, (3r+2), i keep getting 305 instead of 147?! Isn't d=3, a=17 and n=10?! What am i doing wrong??
n = 6 , not 10
13. (Original post by *girlie*)
Wait a second...what do you do if the number above sigma is something like 40?! You dont just add them all up, do you?!
i think its very unlikely that they will ask to go up to 40, that would be stupid.
14. (Original post by Cortez)
i think its very unlikely that they will ask to go up to 40, that would be stupid.
I don't think so, you just put it in the formula, thats whats its' there for!

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