Question

Use the summation formulas to rewrite the expression without the summation notation. summation_k = 1^n 12k(k - 1) / n^3 Use the result to find the sums for n = 10, n= 100, n= 1000 n= 10,000.

Answer 1

Answer:

$Sum=4*[\frac{n^2-1}{n^2}]$

For n=10:

Sum=3.96

For n=100:

Sum=3.9996

For n=1000:

Sum=3.999996

For n= 10000:

Sum=3.99999996

Step-by-step explanation:

Formula:

$\sum_{k=1}^n \frac{12k(k-1)}{n^3}$

Rearranging the above formula:

$\sum_{k=1}^n \frac{12}{n^3}*k(k-1)\\\sum_{k=1}^n \frac{12}{n^3}*(k^2-k)$            Eq (1)

According to summation formula:

$\sum_{k=1}^n\ k= \frac{n(n+1)}{2}\\ \sum_{k=1}^n\ k^2= \frac{n(n+1)(2n+1)}{6}$

Putt these in Eq (1), and we will get:

$=\frac{12}{n^3}[\frac{n(n+1)(2n+1)}{6}-\frac{n(n+1)}{2}]\\Taking\ n\ as\ common\\=n*\frac{12}{n^3}[\frac{(n+1)(2n+1)}{6}-\frac{(n+1)}{2}] \\=\frac{12}{n^2}*[\frac{(n+1)(2n+1)}{6}]-\frac{12}{n^2}*[\frac{(n+1)}{2}] \\=\frac{2*(n+1)(2n+1)}{n^2}-\frac{6(n+1)}{n^2}$

Taking $2(n+1)$ as common:

$=2(n+1)*\frac{2n+1-3}{n^2} \\=(2n+2)*\frac{2n-2}{n^2}\\=\frac{4n^2-4}{n^2}$

After more simplifying,

$Sum=4*[\frac{n^2-1}{n^2}]$

Now ,for n=10:

$Sum=4[\frac{(10^{2})-1}{10^{2}}]\\Sum=3.96$

For n=100:

$Sum=4[\frac{(100^{2})-1}{100^{2}}]\\Sum=3.9996$

For n=1000

$Sum=4[\frac{(1000^{2})-1}{1000^{2}}]\\Sum=3.999996$

For n=10000:

$Sum=4[\frac{(10000^{2})-1}{10000^{2}}]\\Sum=3.99999996$