anonymous
  • anonymous
HELP! with summation notation. will medal! (see attachment)
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
anonymous
  • anonymous
please please help with the first one
anonymous
  • anonymous
Do you need to find the sum?

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anonymous
  • anonymous
yes @Nixy
anonymous
  • 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
anonymous
  • anonymous
Do you still need help?
anonymous
  • anonymous
Now we use the formula \( \huge S_n = \frac{n}{2}(a_1+a_n) \) Now just plugin our numbers \( \huge S_{50} = \frac{50}{2}(10+157) \) \( \huge S_{50} = 4175\)
anonymous
  • anonymous
is that the answer?? thats how you do it? @Nixy
anonymous
  • anonymous
That is all the steps and the answer is 4175
anonymous
  • anonymous
could you also help with 7? then ill do 8 and 9 by my self @Nixy
anonymous
  • anonymous
@pooja195 please help with 7!!??
anonymous
  • anonymous
Number 7 looks like the Sum of an Infinite Geometric Series
anonymous
  • anonymous
if i do it will you check it?
anonymous
  • anonymous
actually i have number 7 done but I dont understand 8
anonymous
  • anonymous
i dont get how there is no number above the "e"
anonymous
  • anonymous
@Nixy
anonymous
  • 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.
anonymous
  • anonymous
is this number 8? @Nixy

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