Question

Find the largest integer value for the upper limit k

if the two series are equal...

Answer 1

\[\sum_{n}^{k}=1(n ^{2}+n)=\sum_{n}^{k}=1[(n-1)(n-2)(n-3)(n-4)(n-5)+n ^{2}+n]\]

Answer 2

you think you could help me out with this problem?

Answer 3

any tips, biddy?

Answer 4

yeah, blacklabel :)

Answer 5

any tips or hints?

Answer 6

ahh

Answer 7

I just don't really know where to start with this problem :(

Answer 8

do you think anyone else could help me?

Answer 9

hello saeddiscover, do you think you could help me out with this problem?

Answer 10

That sounds interesting! I will try ...

Answer 11

thank you :)

Answer 12

what is lower limit of n ?? It is needed to evaluate the summation...

Answer 13

hmm

Answer 14

would the lower limit be 1?

Answer 15

It may be. Are you given any lower limit for n? it should be known ..

Answer 16

I'm not given it

Answer 17

unless the problem has a typo or something

Answer 18

supposing it was 1...

Answer 19

the lower limit?

Answer 20

Yep

Answer 21

how would I write it out as?

Answer 22

I think it is an unwise question, it has no solution.

Answer 23

but if the lower limit was 1, how would I solve it?

Answer 24

what is equal sign (=) for in right hand side of your equation?

Answer 25

my professor wrote the problem during the weekend; do you think he made a typo?

Answer 26

Oops! I meant left-hand side of your equation.

Answer 27

phi!!!!!! do you think you could help me out with this problem :)?

Answer 28

out of curiosity, are you sure that is a one in
1(n^2 + n)

people don't put in leading ones... so maybe it's a typo or means something else ?

Answer 29

oh, I think it is part of n=1 (the lower limit)
\[\sum_{n=1}^{k}(n^2+n)\]

Answer 30

so that is wat the lower limit is?

Answer 31

my bad

Answer 32

yes, the =1 part belongs below the sigma (summation sign)

Answer 33

so, what do I do next? The reason I have so much trouble with these problems

is because I have a Professor that has trouble speaking English and trouble

expressing his ideas :(.

Answer 34

so anymore tips, phi?

Answer 35

if we write it like this
\[ \sum_{n=1}^{k}(n^2+n)= \sum_{n=1}^{k}(n^2+n) + \sum_{n=1}^{k}(n-1)(n-2)(n-3)(n-4)(n-5)\]

it seems the complicated sum must be zero :
\[ \sum_{n=1}^{k}(n-1)(n-2)(n-3)(n-4)(n-5)=0\]

Answer 36

that suggests that as you increase n from 1 to 2 to 3, you keep getting 0
until n is bigger than 5
so we can go from n=1 to n=5 and get a sum of zero

Answer 37

wait so

Answer 38

(1-1)(2-2)(3-3)(4-4)(5-5)?

Answer 39

But there are no (n-6)(n-7)(n-8)...

Answer 40

so is the upper limit 5?

Answer 41

the professor seems to be very crazy...

Answer 42

the sum
\[ \sum_{n=1}^{k}(n-1)(n-2)(n-3)(n-4)(n-5) \]
means, add up
(1-1)(1-2)(1-3)(1-4)(1-5) + (2-1)(2-2)(2-3)(2-4)(2-5) + and so on

Answer 43

phi, sorry about earlier, apparently open study had to reboot

Answer 44

notice that the first term in the summation is
(1-1)(1-2)(1-3)(1-4)(1-5)
and (1-1) is 0, so this turns into 0

the second term (2-1)(2-2)(2-3)(2-4)(2-5) is also 0 (because you have (2-2) in it)

each term will be zero...
when you get to n=5 you have the term
(5-1)(5-2)(5-3)(5-4)(5-5)
and again you get 0 (because you are multiplying by (5-5)=0

if you go to n=6, that term is
(6-1)(6-2)(6-3)(6-4)(6-5)
and this is NOT zero. so the biggest k can be (and still get a sum of zero) is k=5

Answer 45

wow

Answer 46

incredible

Answer 47

so phi

Answer 48

is that the solution to the problem?

Answer 49

how bout the n squared+n part?

Answer 50

to summarize
\[ \sum_{n=1}^{k}(n^2+n)= \sum_{n=1}^{k}(n^2+n) +(n-1)(n-2)(n-3)(n-4)(n-5) \\
\sum_{n=1}^{k}(n^2+n)= \sum_{n=1}^{k}(n^2+n) + \sum_{n=1}^{k}(n-1)(n-2)(n-3)(n-4)(n-5) \]
subtract \(\sum_{n=1}^{k}(n^2+n)\) from both sides to get
\[ 0=\sum_{n=1}^{k}(n-1)(n-2)(n-3)(n-4)(n-5) \]
you can go up to n=5 and get a sum of zero
\[ (1-1)(1-2)(1-3)(1-4)(1-5) + (2-1)(2-2)(2-3)(2-4)(2-5) +...+\\
(5-1)(5-2)(5-3)(5-4)(5-5) =0\]

Answer 51

if you go to n=6, that term is
(6-1)(6-2)(6-3)(6-4)(6-5)
and this is NOT zero. so the biggest k can be (and still get a sum of zero) is k=5

Answer 52

ahh

Answer 53

you are such a big help

Answer 54

I give you an S+