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Mathematics 14 Online
OpenStudy (anonymous):

write the series with summation notation. 2+5+8+11+ . . .

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

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 ?

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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