Question

How do you find the partial sum of $\sum\limits_{n = 1}^{100} {2n}$ ?

Answer 1

In this question we have to find the partial sum for the given notation, this can be done by using the method of Gauss and the Gauss formula allows us to add together the first n positive integers, and this is given by $\sum\limits_{i = 1}^n i = \dfrac{{n\left( {n + 1} \right)}}{2}$ where $n$ is the integer and by substituting the values from the question in the formula and further simplification we will get the required answer.

Complete step by step answer:
Given expression is $\sum\limits_{n = 1}^{100} {2n}$ , we have to find the partial sum,
We now use the Gauss method which is given by $\sum\limits_{i = 1}^n i = \dfrac{{n\left( {n + 1} \right)}}{2}$ ,
Here substitute the values from the given expression i.e., $n = 100$ and $i = n$ ,
Now rewrite given expression as,
$\Rightarrow \sum\limits_{n = 1}^{100} {2n = } 2\sum\limits_{n = 1}^{100} n$ ,
Now applying the Gauss formula we get,
$\Rightarrow \sum\limits_{n = 1}^{100} {2n = } 2\dfrac{{n\left( {n + 1} \right)}}{2}$ ,
Now we know that $n = 100$ , substituting the value of $n$ we get,
$\Rightarrow \sum\limits_{n = 1}^{100} {2n = } 2 \times \dfrac{{100\left( {100 + 1} \right)}}{2}$ ,
Now simplifying by eliminating numbers in numerator and denominator, we get,
$\Rightarrow \sum\limits_{n = 1}^{100} {2n = } 100\left( {101} \right)$ ,
Now multiplying the numbers we get,
$\Rightarrow \sum\limits_{n = 1}^{100} {2n = } 10100$ ,
So, the partial sum is equal to 10100.

The partial sum for the given notation $\sum\limits_{n = 1}^{100} {2n}$ will be equal to 10100.

Note: A Partial Sum is a Sum of Part of a Sequence. A Sequence is a set of things (usually numbers) that are in order. Partial Sums are often written using $\sum {}$ to mean "add them all up":
This symbol (called Sigma) $\sum {}$ means "sum up". Here some of useful formulas in $\sum {}$ :
$\sum\limits_{k = 1}^n 1 = n$ , summing 1 equal to n,
$\sum\limits_{k = 1}^n c = cn$, summing constants c equal to c times n,
$\sum\limits_{k = 1}^n k = \dfrac{{n\left( {n + 1} \right)}}{2}$, summing of k terms,
$\sum\limits_{k = 1}^n {{k^2}} = \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}$ , summing of ${k^2}$ terms,
$\sum\limits_{k = 1}^n {{k^3}} = {\left( {\dfrac{{n\left( {n + 1} \right)}}{2}} \right)^2}$, summing o ${k^3}$ terms,
$\sum\limits_{k = 1}^n {\left( {2k - 1} \right)} = {n^2}$, summing of odd numbers.
$\sum\limits_{k = 1}^n {\left( {2k - 1} \right)} = {n^2}$ .