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Mathematics 19 Online
OpenStudy (anonymous):

HELP! with summation notation. will medal! (see attachment)

OpenStudy (anonymous):

OpenStudy (anonymous):

please please help with the first one

OpenStudy (anonymous):

Do you need to find the sum?

OpenStudy (anonymous):

yes @Nixy

OpenStudy (anonymous):

The first one is arithmetic so we can do the following First find the first and last terms by putting n= 1 and n = 50 \( \huge a_1 = 3(1) + 7 = 10\) 10 is your first. Now find the last \( \huge a_1 = 3(50) + 7 = 157\) Ok, next part coming. I am doing number 6

OpenStudy (anonymous):

Do you still need help?

OpenStudy (anonymous):

Now we use the formula \( \huge S_n = \frac{n}{2}(a_1+a_n) \) Now just plugin our numbers \( \huge S_{50} = \frac{50}{2}(10+157) \) \( \huge S_{50} = 4175\)

OpenStudy (anonymous):

is that the answer?? thats how you do it? @Nixy

OpenStudy (anonymous):

That is all the steps and the answer is 4175

OpenStudy (anonymous):

could you also help with 7? then ill do 8 and 9 by my self @Nixy

OpenStudy (anonymous):

@pooja195 please help with 7!!??

OpenStudy (anonymous):

Number 7 looks like the Sum of an Infinite Geometric Series

OpenStudy (anonymous):

if i do it will you check it?

OpenStudy (anonymous):

actually i have number 7 done but I dont understand 8

OpenStudy (anonymous):

i dont get how there is no number above the "e"

OpenStudy (anonymous):

@Nixy

OpenStudy (anonymous):

\( \huge \sum_{n=1}^{12} 2*4^n \) First you need to rewrite the series so it is in the form of \( \huge \sum_{n=1}^{12} a_1*r^{n-1} \) \( \huge \sum_{n=1}^{12} 8*4^{n-1} \) Now identify a_1 and r ? If |r| < 1 the series is converges and we need to find the sum if not it, there is no need to find the sum Sorry I am at work and that is why it is taking me a little bit of time.

OpenStudy (anonymous):

is this number 8? @Nixy

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