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Mathematics 21 Online
OpenStudy (anonymous):

PLEASE HELP!! MEDAL WILL BE GIVEN. ALSO FAN :)

OpenStudy (anonymous):

\[\sum_{3}^{5} (k^2 - 10)\]

OpenStudy (anonymous):

Find the sum of the sequence.

OpenStudy (anonymous):

@jim_thompson5910 @mathslover @mathsmind @satellite73 @wio @AravindG @dumbcow

OpenStudy (anonymous):

well u can use the formula for solving this, note u may split ur summation into sum of k^2 and sum of -10

OpenStudy (anonymous):

ok i did that

OpenStudy (anonymous):

i will finish the calculus guy and come and solve ur problem

OpenStudy (anonymous):

please do as soon as u can? anyone else can help me right now??

OpenStudy (anonymous):

@jim_thompson5910 @mathslover @mathsmind @satellite73 @wio @AravindG @dumbcow

OpenStudy (anonymous):

\[\sum_{2}^{5}k^2-\sum_{2}^{5}10\]

OpenStudy (anonymous):

okaii got you, whats next

OpenStudy (anonymous):

@mathslover @mathsmind

OpenStudy (anonymous):

???

OpenStudy (anonymous):

wait i am replying to 002830948028498298409238 msgs hehehehe

OpenStudy (anonymous):

and i am coding with php be patient please

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

PLEASE SOMEONE HELP ME :(

OpenStudy (anonymous):

@ghazi

OpenStudy (anonymous):

@jim_thompson5910

OpenStudy (anonymous):

@karatechopper

OpenStudy (anonymous):

ill help u if u help meh

OpenStudy (anonymous):

\[=\frac{1}{6}k(k+1)(2k+1)\]

OpenStudy (anonymous):

how did u get that?

OpenStudy (anonymous):

this is the formula for summing K^2

OpenStudy (anonymous):

u can prove that using induction

OpenStudy (anonymous):

and the sum of 10 would be 10k

OpenStudy (anonymous):

now there is a trick in ur question

OpenStudy (anonymous):

yeah?

OpenStudy (anonymous):

k should start from 1, so u need to shift ur sum by 2 units do u get that?

OpenStudy (anonymous):

so the whole summation must be shifted this is the only trick of ur question

OpenStudy (anonymous):

so how do i do that?

OpenStudy (anonymous):

the way i do it!

OpenStudy (anonymous):

how?

OpenStudy (anonymous):

@dumbcow please help?

OpenStudy (anonymous):

@mathsmind

jimthompson5910 (jim_thompson5910):

when k = 3 k^2 - 10 =3^2 - 10 = 9 - 10 = -1 so when k = 3, the term is -1

jimthompson5910 (jim_thompson5910):

when k = 4 k^2 - 10 =4^2 - 10 = 16 - 10 = 6 when k = 4, the term is 6

jimthompson5910 (jim_thompson5910):

when k = 5 k^2 - 10 =5^2 - 10 = 25 - 10 = 15 so when k = 5, the term is 15

jimthompson5910 (jim_thompson5910):

add up the terms: -1 + 6 + 15 = 20

jimthompson5910 (jim_thompson5910):

therefore, \[\Large \sum_{3}^{5} (k^2 - 10) = 20\]

OpenStudy (anonymous):

why did u stop when k = 5

OpenStudy (anonymous):

the answer is 20

OpenStudy (anonymous):

because the question asks u to stop at 5

jimthompson5910 (jim_thompson5910):

\[\Large \sum_{3}^{5} (k^2 - 10)\] means you add the terms for k = 3, k = 4 and k = 5 and that's it

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