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OpenStudy (anonymous):
write the series with summation notation.
2+5+8+11+ . . .
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hartnn (hartnn):
so you need to find n'th term first
do u notice any pattern ?
OpenStudy (anonymous):
THERE IS A CERTAIN PATTEN..OF 2+(2+3)+(2+3+3)+(2+3+3+3)....
WRITING IN SUMMATION IT WILL BE \[\sum_{1}^{N}2+\sum_{1}^{N-1}3\]
hartnn (hartnn):
thats not totally correct...first of all, its an infinite series....
OpenStudy (anonymous):
DOES THE QUESTION GIVES ANY INTEL ABOUT THAT? WELL THEN I AM CORRECT IF ITS FOR N TH TERMS..
OpenStudy (anonymous):
The sum doesn't work
you can test it out yourself
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OpenStudy (anonymous):
the pattern is +3
hartnn (hartnn):
correct!
so we can say the common difference between the terms = d = +3
hartnn (hartnn):
first term = a = 2
now use the formula for n'th term of this arithmetic series as
\(\large a_n = a +(n-1)d\)
hartnn (hartnn):
then you can put that a_n in sum notation as
\(\Large \sum \limits_1^\infty a_n\)
hartnn (hartnn):
try it ?
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hartnn (hartnn):
\(a_n = 2+(n-1)\times 3= 3n-1 \\ so, \sum \limits_1^\infty (3n-1)\)
thats it!
OpenStudy (anonymous):
that's the answer?
hartnn (hartnn):
yes
OpenStudy (anonymous):
is it correct? I have a picky teacher :( lol
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