Question

How do you use the summation formulas to rewrite the expression $\sum {\dfrac{{4{i^2}\left( {i - 1} \right)}}{{{n^4}}}}$ as k=1 to n without the summation notation and then use the result to find the sum for n=10,100,1000and 10000?

Answer 1

Hint : The above question is based on the concept of summation notation. The main approach towards solving the expression is by splitting the terms in the numerator then applying summation property on individual terms to get the general solution.

Complete step-by-step answer :
Summation notation allows us to write a long sum in a single expression. $\sum {}$ denotes sigma which is a summation symbol. The range 1 to n denotes the sum of n numbers. If n=10 then we can say it is the sum of 10 numbers.
So on solving the above expression we get,
Let

$$\dfrac{{4{i^2}\left( {i - 1} \right)}}{{{n^4}}} \;$$

So now applying the summation on individual terms,
{S_n} = \dfrac{4}{{{n^4}}}\sum\limits_{i = 1}^n {\left( {{i^3} - {i^2}} \right)} = \dfrac{4}{{{n^4}}}\left\\{ {\sum\limits_{i = 1}^n {{i^3}} - \sum\limits_{i = 1}^n {{i^2}} } \right\\}
According to the standard results,
$\sum\limits_{r = 1}^n {{r^2}} = \dfrac{1}{6}n(n + 1)(2n + 1)$ and $\sum\limits_{r = 1}^n {{r^3} = \dfrac{1}{4}{n^2}{{\left( {n + 1} \right)}^2}}$
Therefore, by substituting it we get,

\Rightarrow {S_n} = \dfrac{4}{{{n^4}}}\left\\{ {\dfrac{1}{4}{n^2}{{\left( {n + 1} \right)}^2} - \dfrac{1}{6}n\left( {n + 1} \right)\left( {2n + 1} \right)} \right\\} \\\ \Rightarrow {S_n} = \dfrac{4}{{{n^4}}}\left( {\dfrac{{n\left( {n + 1} \right)}}{{12}}} \right)\left\\{ {3n(n + 1) - 2(2n + 1)} \right\\} \\\ \Rightarrow {S_n} = \dfrac{{\left( {n + 1} \right)}}{{3{n^3}}}\left( {3{n^2} - n - 2} \right) \;

Now by substituting values of n we get,
When n=10, then ${S_n}$ will be 1.05600
When n=100 ,then ${S_n}$ will be 1.0065
When n=1000, then ${S_n}$ will be 1.00060
When n=10000, then ${S_n}$ will be 1.000065
Thus, with the help of the above general solution we get the values of the sum of first 10 ,100 ,1000, etc terms.

Note : An important thing to note is that when n tends to infinity $n \to \infty$ then ${S_n} \to 1$ .This can be proved by substituting the value of n. Since n is in the denominator therefore $\dfrac{1}{\infty } \to 0$ because infinity is in the denominator it always tends to zero. Hence all the terms when added is equal to one