Mathematics
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OpenStudy (anonymous):
PLEASE HELP!!
MEDAL WILL BE GIVEN. ALSO FAN :)
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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
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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
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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
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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
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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
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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?
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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
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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
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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