# 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?

##### 1 Answer

#### Explanation:

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:

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

**What happens as #n rarr oo#?**

[ NB As an additional task we could possibly conclude that as

Now,

# :. 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!