How do you use the summation formulas to rewrite the expression #Sigma (4j+3)/n^2# as j=1 to n without the summation notation and then use the result to find the sum for n=10, 100, 1000, and 10000?

Answer 1

# sum_(j=1)^n 1/n^2(4j+3) = (2n + 5)/n #

Let # S_n = sum_(j=1)^n 1/n^2(4j+3) #
# :. S_n = 1/n^2sum_(j=1)^n (4j+3) #
# :. S_n = 1/n^2{sum_(j=1)^n (4j) + sum_(j=1)^n (3)} #
# :. S_n = 1/n^2{4 sum_(j=1)^n (j) + sum_(j=1)^n (3)} #

And using the standard results:
# sum_(r=1)^n r = 1/2n(n+1) #

We have;

# S_n = 1/n^2{4 1/2n(n+1) + 3n} #
# :. S_n = 1/n^2(2n(n+1) + 3n) #
# :. S_n = 1/n^2(2n^2+2n + 3n) #
# :. S_n = 1/n^2(2n^2 + 5n) #
# :. S_n = 1/n^2(2n + 5)n #
# :. S_n = (2n + 5)/n #

And this has been calculated using Excel for #n=10, 100, 1000, 10000#

What happens as #n rarr oo#?

[ NB As an additional task we could possibly conclude that as #n rarr oo# then #S_n rarr 2#; This is probably the conclusion of this question]

Now, # S_n = (2n + 5)/n #

# :. S_n = 2n/n + 5/n #
# :. S_n = 2 + 5/n #

And so,

# lim_(n rarr oo) S_n = lim_(n rarr oo) (2 + 5/n) #
# :. lim_(n rarr oo) S_n = lim_(n rarr oo) (2) + lim_(n rarr oo)(5/n) #
# :. lim_(n rarr oo) S_n = 2 + 5 lim_(n rarr oo)(1/n) #
# :. lim_(n rarr oo) S_n = 2 #, #" as "(lim_(n rarr oo)(1/n) =0)#
Which confirms our assumption!

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Answer 2

To rewrite the expression (\sum_{j=1}^{n} \frac{4j+3}{n^2}) without summation notation, first, express (4j+3) as a separate summation:

[ \sum_{j=1}^{n} \frac{4j+3}{n^2} = \frac{1}{n^2} \sum_{j=1}^{n} (4j+3) ]

Next, use the summation formulas for arithmetic series to evaluate (\sum_{j=1}^{n} (4j+3)) which is given by ( \frac{n}{2} [2a + (n-1)d] ), where (a) is the first term, (d) is the common difference, and (n) is the number of terms.

In this case, (a = 4(1) + 3 = 7), (d = 4), and (n) varies depending on the value given.

For (n=10), (n=100), (n=1000), and (n=10000), we have:

[ n=10: \frac{10}{2}[2(7) + (10-1)(4)] = 55 ] [ n=100: \frac{100}{2}[2(7) + (100-1)(4)] = 5050 ] [ n=1000: \frac{1000}{2}[2(7) + (1000-1)(4)] = 500500 ] [ n=10000: \frac{10000}{2}[2(7) + (10000-1)(4)] = 50005000 ]

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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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