Answered: "Use the summation formulas to rewrite the expression..."

Harold Prince

Answered question

2023-01-17

Use the summation formulas to rewrite the expression without the summation notation. Use the result to find the sums for

n = 10, 100, 1000 and 10, 000

\sum k = 1 n \frac{6 k (k - 1)}{n^{3}}

possetzvjm

Beginner2023-01-18Added 7 answers

Step-1 : Summation of series:
Consider the expression

\sum k = 1 n \frac{6 k (k - 1)}{n^{3}}

.
Since the summation runs over

k

, take the constant

\frac{6}{n^{3}}

out of summation:

\frac{6}{n^{3}} \sum k = 1 n k (k - 1)

Simplify the expression:

\frac{6}{n^{3}} \sum k = 1 n k (k - 1) = \frac{6}{n^{3}} \sum k = 1 n k^{2} - k = \frac{6}{n^{3}} (\sum k = 1 n k^{2} - \sum k = 1 n k)

Then, use the standard result

\sum k = 1 n k^{2} = \frac{n (n + 1) (2 n + 1)}{6} and \sum k = 1 n k = \frac{n (n + 1)}{2}

in the above expression:

\frac{6}{n^{3}} (\sum k = 1 n k^{2} - \sum k = 1 n k) = \frac{6}{n^{3}} (\frac{n (n + 1) (2 n + 1)}{6} - \frac{n (n + 1)}{2})

Further simplify the expression:

\frac{6}{n^{3}} (\sum k = 1 n k^{2} - \sum k = 1 n k) = \frac{6}{n^{3}} (\frac{n (n + 1) (2 n + 1)}{6} - \frac{n (n + 1)}{2}) = \frac{6}{n^{2}} (\frac{(n + 1) (2 n + 1)}{6} - \frac{(n + 1)}{2}) = \frac{(n + 1) (2 n + 1)}{n^{2}} - \frac{3 (n + 1)}{n^{2}}

Therefore, the rewritten form of the expression

\sum k = 1 n \frac{6 k (k - 1)}{n^{3}}

without the summation notation is

\frac{(n + 1) (2 n + 1)}{n^{2}} - \frac{3 (n + 1)}{n^{2}}

.
Next, find the sum for

n = 10, 100, 1000 and 10, 000

by substituting the respective values in the rewritten expression

\frac{(n + 1) (2 n + 1)}{n^{2}} - \frac{3 (n + 1)}{n^{2}}

:
Step-2: Summation for

n = 10

\frac{(n + 1) (2 n + 1)}{n^{2}} - \frac{3 (n + 1)}{n^{2}} = \frac{(10 + 1) (2 \cdot 10 + 1)}{{(10)}^{2}} - \frac{3 (10 + 1)}{{(10)}^{2}} = \frac{11 \cdot 21}{100} - \frac{3 \cdot 11}{100} = 2.31 - 0.33 = 1.98

Step-3: Summation for

n = 100

\frac{(n + 1) (2 n + 1)}{n^{2}} - \frac{3 (n + 1)}{n^{2}} = \frac{(100 + 1) (2 \cdot 100 + 1)}{{(100)}^{2}} - \frac{3 (100 + 1)}{{(100)}^{2}} = 2.0301 - 0.0303 = 1.9998

Step-4 : Summation for

n = 1000

\frac{(n + 1) (2 n + 1)}{n^{2}} - \frac{3 (n + 1)}{n^{2}} = \frac{(1000 + 1) (2 \cdot 1000 + 1)}{{(1000)}^{2}} - \frac{3 (1000 + 1)}{{(1000)}^{2}} = 2.003001 - 0.003003 = 1.999998

Step-5 : Summation for

n = 10, 000

\frac{(n + 1) (2 n + 1)}{n^{2}} - \frac{3 (n + 1)}{n^{2}} = \frac{(10, 000 + 1) (2 \cdot 10, 000 + 1)}{{(10, 000)}^{2}} - \frac{3 (10, 000 + 1)}{{(10, 000)}^{2}} = 2.00030001 - 0.00030003 = 1.99999998

Therefore the sum for

n = 10, 100, 1000 and 10, 000

i is

1.98, 1.9998, 1.999998 and 1.99999998

respectively.

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