Question

How do you use the summation formula to rewrite the expression $\sum{\dfrac{2i+1}{{{n}^{2}}}}$ as i = 1 to n without the summation notation then use the result to find the sum for n = 10, 100, 1000, 10000?

Answer 1

Hint: Now we are given with the summation term. We will first split the summation to each term and take the constants out of the summation. Then we will use the known summation formulas to write the values of $\sum{i}$ and $\sum{1}$ . Hence using this we will simplify the expression obtained and hence find the general expression for the summation.

Complete step by step solution:
Now consider the given summation $\sum{\dfrac{2i+1}{{{n}^{2}}}}$
Now since n is a constant we can easily take ${{n}^{2}}$ out of the summation Hence we get,
\[\Rightarrow \sum{\dfrac{2i+1}{{{n}^{2}}}}=\dfrac{1}{{{n}^{2}}}\sum{\left( 2i+1 \right)}\]
Now using summation on each term we get,
\[\Rightarrow \sum{\dfrac{2i+1}{{{n}^{2}}}}=\dfrac{1}{{{n}^{2}}}\left( \sum{\left( 2i \right)+\sum{1}} \right)\]
Now again we can take 2 out of summation. Hence we get,
\[\Rightarrow \sum{\dfrac{2i+1}{{{n}^{2}}}}=\dfrac{1}{{{n}^{2}}}\left( 2\sum{i+\sum{1}} \right)\]
Now we know by summation formulas that \[\sum\limits_{I=1}^{n}{i}=\dfrac{n\left( n+1 \right)}{2}\] and \[\sum\limits_{1}^{n}{1}=n\]
Hence using this we get,
\[\Rightarrow \sum{\dfrac{2i+1}{{{n}^{2}}}}=\dfrac{1}{{{n}^{2}}}\left( 2\left( \dfrac{n\left( n+1 \right)}{2} \right)+n \right)\]
Cancelling 2 from numerator and denominator we get,
\[\Rightarrow \sum{\dfrac{2i+1}{{{n}^{2}}}}=\dfrac{1}{{{n}^{2}}}\left( n\left( n+1 \right)+n \right)\]
Now taking n common from the brackets and cancelling it with the help of the denominator we get,
\[\begin{align}
  & \Rightarrow \sum{\dfrac{2i+1}{{{n}^{2}}}}=\dfrac{1}{n}\left( \left( n+1 \right)+1 \right) \\
 & \Rightarrow \sum{\dfrac{2i+1}{{{n}^{2}}}}=\dfrac{n+2}{n} \\
\end{align}\]
Now let us check the value of summation for n = 10, 100, 1000, 10000.
Hence we have,
For n = 10 $\sum{\dfrac{2i+1}{{{n}^{2}}}}=\dfrac{12}{10}$
For n = 100 \[\sum{\dfrac{2i+1}{{{n}^{2}}}=}\dfrac{102}{100}\]
For n = 1000, \[\sum{\dfrac{2i+1}{{{n}^{2}}}=}\dfrac{1002}{1000}\]
And for n = 10000, \[\sum{\dfrac{2i+1}{{{n}^{2}}}=}\dfrac{10002}{10000}\]

Note:
Now note that the value of $\sum\limits_{1}^{n}{i}$ and $\sum\limits_{1}^{n}{1}$ can be easily be found by using the concept of Arithmetic progression. We know that the sum of Arithmetic progression is given by ${{S}_{n}}=\dfrac{n\left( 2a+\left( n-1 \right)d \right)}{2}$ . Now we will substitute appropriate values of a and d to find the sum.