Verified answer

Find 6 ∑ n=1 5(-4)^n-1

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The sum of 5(-4)^n-1 form n = 1 to n = 6 is

5.( - 4)^(1 - 1) + 5.( - 4)^(2 - 1) + 5.( - 4)^(3 - 1) + 5.( - 4)^(4 - 1) + 5.( - 4)^(5 - 1) + 5.( - 4)^(6 - 1)

= 5.[ ( - 4)^(1 - 1) + ( - 4)^(2 - 1) + ( - 4)^(3 - 1) + ( - 4)^(4 - 1) + ( - 4)^(5 - 1) + ( - 4)^(6 - 1) ]

= 5.[ ( - 4)^0 + ( - 4)^1 + ( - 4)^2 + ( - 4)^3 + ( - 4)^4 + ( - 4)^5 ]

= 5.(1 - 4 + 16 - 64 + 256 - 1024) --------- ( - 4)^0 = 1 and ( - 4)^1 = - 4

= 5.( - 819)

= - 4095

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Hope this helps!

Think of it like a function of x f(x) where you plug in x values every where they show up in the formula, except with sigma notation it makes the formula by following specific rules, rule one, each segment of the formula is built using the argument, in this case the argument is 5(-4)^n-1 i will assume that -1 is in the exponent... Rule two plug in increasing values of n until you reach the end of the formula

Also im going to factor out 5

....

5( (-4)^n-1 + (-4)^n-1 + •••)

5( (-4)^1-1+(-4)^2-1+ (-4)^3-1+•••)

5(-4^0 + -4^1 + (-4)^2 +•••)

5(1-4+8+•••)

S = 5*[1+(-4)^1 + (-4)^2 +---- + (-4)^5]. Using formula for Geometric series, we get

S = 5*[{1-(-4)^6}/(1-(-4)] = 5*(-4095/5) = -4095