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FP1 Standard result for r

What's the difference between the result

∑r=1nr=n2(n+1)\displaystyle \sum_{r=1}^n r = \dfrac{n}{2} (n+1)

and the result

∑r=1nb=nb\displaystyle \sum_{r=1}^n b = nb
(edited 7 years ago)
Reply 1
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B_9710
18
Original post by huiop
What's the difference between the result

∑r=1nr=n2(n+1)\displaystyle \sum_{r=1}^n r = \dfrac{n}{2} (n+1)
and the result

∑r=1nb=nb\displaystyle \sum_{r=1}^n b = nb


The first one is 1+2+3+...+n 1+2+3+...+n while the first one b+b+b+b+...+b⏟n \underbrace{b+b+b+b+...+b}_{n} .
(edited 7 years ago)
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huiop
OP
1
Original post by B_9710
The first one is 1+2+3+...+n 1+2+3+...+n while the first one b+b+b+b+...+b b+b+b+b+...+b where there are n b's.


so how would i know which one to use in a question which says

show that

∑r=1n(2r−1)2=n3(4n2−1)\displaystyle \sum_{r=1}^n (2r-1)^2 = \dfrac{n}{3} (4n^2 -1)

so far i got to this 4∑r=1nr2−4∑r=1nr+∑r=1n1\displaystyle 4\sum_{r=1}^n r^2 -4\sum_{r=1}^n r +\sum_{r=1}^n 1

then i went to this 4×n6(2n+1)(n+1)−4n+1\dfrac{4\times n}{6} (2n+1)(n+1)-4n+1

so that 4n is wrong so how would i know to use standard result for r instead of b?
(edited 7 years ago)
Original post by huiop
so how would i know which one to use in a question which says

show that

∑r=1n(2r−1)2=n3(4n2−1)\displaystyle \sum_{r=1}^n (2r-1)^2 = \dfrac{n}{3} (4n^2 -1)

so far i got to this 4∑r=1nr2−4∑r=1nr+∑r=1n1\displaystyle 4\sum_{r=1}^n r^2 -4\sum_{r=1}^n r +\sum_{r=1}^n 1

then i went to this 4xn6(2n+1)(n+1)−4n+1\dfrac{4xn}{6} (2n+1)(n+1)-4n+1

so that 4n is wrong so how would i know to use standard result for r instead of b?


Where does the xx come from?

4∑r=1nr2=4[16n(2n+1)(n+1)]\displaystyle 4\sum_{r=1}^n r^2 =4[\frac{1}{6}n(2n+1)(n+1)]
−4∑r=1nr=−4[12n(n+1)]\displaystyle -4\sum_{r=1}^n r =-4[\frac{1}{2}n(n+1)]
∑r=1n1=n\displaystyle \sum_{r=1}^n 1 =n
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huiop
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Original post by RDKGames
Where does the xx come from?

4∑r=1nr2=4[16n(2n+1)(n+1)]\displaystyle 4\sum_{r=1}^n r^2 =4[\frac{1}{6}n(2n+1)(n+1)]
−4∑r=1nr=−4[12n(n+1)]\displaystyle -4\sum_{r=1}^n r =-4[\frac{1}{2}n(n+1)]
∑r=1n1=n\displaystyle \sum_{r=1}^n 1 =n


whoops typing error i thought i could use x as a substitute for \times
Original post by huiop
whoops typing error i thought i could use x as a substitute for \times


As for the difference between rr and bb;

If it's r, then you are adding 1+2+3+4+5+...+n because you are summing up the integers from r=1 up to r=n.

If it's a constant (like b), then you are adding b+b+b+b+b+...+n because you are summing up a constant b, in a series of b, b, b, b, b, b, .... where every single term is b. So r=1 will be b, and r=2 will be b, and r=3 and so on up to r=n. Hence where nb comes from. As you are summing b an n amount of times.
(edited 7 years ago)
Reply 6
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huiop
OP
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Original post by RDKGames
As for the difference between rr and bb;

If it's r, then you are adding 1+2+3+4+5+...+n because you are summing up the integers from r=1 up to r=n.

If it's a constant (like b), then you are adding b+b+b+b+b+...+n because you are summing up a constant b, in a series of b, b, b, b, b, b, .... where every single term is b. So r=1 will be be, and r=2 will be b, and r=3 and so on up to r=n. Hence where nb comes from. As you are summing b an n amount of times.


ok thanks for clarification
Reply 7
Avatar for B_9710
B_9710
18
Original post by huiop
so how would i know which one to use in a question which says

show that

∑r=1n(2r−1)2=n3(4n2−1)\displaystyle \sum_{r=1}^n (2r-1)^2 = \dfrac{n}{3} (4n^2 -1)

so far i got to this 4∑r=1nr2−4∑r=1nr+∑r=1n1\displaystyle 4\sum_{r=1}^n r^2 -4\sum_{r=1}^n r +\sum_{r=1}^n 1

then i went to this 4×n6(2n+1)(n+1)−4n+1\dfrac{4\times n}{6} (2n+1)(n+1)-4n+1

so that 4n is wrong so how would i know to use standard result for r instead of b?


Think about what this notation means. If you have ∑r=1n1=1+1+1+...+1=n \displaystyle \sum_{r=1}^n 1 = 1+1+1+...+1 =n since there are n 1's. Using this notation, the sum is with respect to r r only. So if we have
Unparseable latex formula:

\displaystle \sum_{r=1}^n b^2 = nb^2

not (1/6)n(n+1)(2n+1) (1/6)n(n+1)(2n+1) .

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