How do you find a formula for the sum n terms #Sigma (1+(2i)/n)^3(2/n)# and then find the limit as #n->oo#?

Answer 1

# sum_(i=1)^n(1 + (2i)/n)^3(2/n)= 2/n^2(10n^2+13n+4)#
# lim_(n rarr oo)sum_(i=1)^n(1 + (2i)/n)^3(2/n)= 20#

We will need the standard results: # sum_(r=1)^n r = 1/2n(n+1) # # sum_(r=1)^n r^2 = 1/6n(n+1)(2n+1) # # sum_(r=1)^n r^2 = 1/4n^2(n+1)^2 = (sum_(r=1)^n r)^2 #

Let

# S_n = sum_(i=1)^n(1 + (2i)/n)^3(2/n) # # :. S_n = 2/n sum_(i=1)^n((n + 2i)/n)^3 # # :. S_n = 2/n sum_(i=1)^n1/n^3(n + 2i)^3 # # :. S_n = 2/n^4 sum_(i=1)^n(n + 2i)^3 # # :. S_n = 2/n^4 sum_(i=1)^n(n^3 + 3n^2(2i) + 3n(2i)^2 + (2i)^3) # # :. S_n = 2/n^4 sum_(i=1)^n(n^3 + 6n^2i + 12ni^2 + 8i^3) # # :. S_n = 2/n^4 {sum_(i=1)^n n^3 + sum_(i=1)^n 6n^2i + sum_(i=1)^n 12ni^2 + sum_(i=1)^n 8i^3 }# # :. S_n = 2/n^4 {n^3sum_(i=1)^n 1 + 6n^2sum_(i=1)^n i + 12nsum_(i=1)^n i^2 + 8sum_(i=1)^n i^3 }# # :. S_n = 2/n^4 {n^3 n + 6n^2 1/2n(n+1) + 12n 1/6n(n+1)(2n+1) + 8 1/4n^2(n+1)^2 }# # :. S_n = 2/n^4 {n^4 + 3n^3 (n+1) + 2n^2(n+1)(2n+1) + 2n^2(n+1)^2 }# # :. S_n = 2/n^4 n^2 {n^2 +3n (n+1) + 2(n+1)(2n+1) + 2(n+1)^2 }# # :. S_n = 2/n^2 {n^2 +3n^2+3n + 2(2n^2+3n+1) + 2(n^2+2n+1) }# # :. S_n = 2/n^2{n^2 +3n^2+3n + 4n^2+6n+2 + 2n^2+4n+2 }# # :. S_n = 2/n^2(10n^2+13n+4)#

Hence,

# sum_(i=1)^n(1 + (2i)/n)^3(2/n)= 2/n^2(10n^2+13n+4)#

And so;

# lim_(n rarr oo)S_n = 2/n^2(10n^2+13n+4)# # lim_(n rarr oo)S_n = 20 + 26/n + 8/n^2 # # lim_(n rarr oo)S_n = 20 # as both #1/n rarr 0#, and #1/n^2 rarr 0# as #n rarr oo#

Hence,

# lim_(n rarr oo)sum_(i=1)^n(1 + (2i)/n)^3(2/n)= 20#
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Answer 2

To find a formula for the sum of ( n ) terms of the series ( \sum_{i=1}^{n} \left(1 + \frac{2i}{n}\right)^3 \cdot \frac{2}{n} ), we first rewrite it as ( \frac{2}{n} \sum_{i=1}^{n} \left(1 + \frac{2i}{n}\right)^3 ). Then, we recognize this as a Riemann sum for the function ( f(x) = (1 + 2x)^3 ) over the interval ( [0,1] ) with ( n ) subintervals.

Using the formula for the Riemann sum, we have:

[ \lim_{n \to \infty} \frac{1}{n} \sum_{i=1}^{n} f\left(\frac{i}{n}\right) = \int_{0}^{1} f(x) , dx ]

Substituting ( f(x) = (1 + 2x)^3 ) into the integral:

[ \int_{0}^{1} (1 + 2x)^3 , dx ]

We integrate this function to find the sum of the series.

[ \int_{0}^{1} (1 + 2x)^3 , dx = \left[\frac{1}{8}(1 + 2x)^4\right]_{0}^{1} = \frac{1}{8}(1 + 2)^4 - \frac{1}{8}(1 + 0)^4 ]

[ = \frac{1}{8}(3^4) - \frac{1}{8}(1^4) = \frac{81 - 1}{8} = \frac{80}{8} = 10 ]

Therefore, the formula for the sum of ( n ) terms is ( \boxed{10} ).

To find the limit as ( n ) approaches infinity, we use the formula for the sum of an infinite series, which is simply the limit of the sequence of partial sums. Since we have already found that the sum of ( n ) terms is 10, the limit as ( n ) approaches infinity is also (\boxed{10}).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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