Answer
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Hint: There are two forms possible for the sums, or the series. One is the sigma form and the other is the expanded form. The sigma form is also known as the contracted form of a sum, and is written in the form of \[\sum\limits_{r={{r}_{1}}}^{{{r}_{2}}}{f\left( r \right)}\]. Here $f\left( r \right)$ is a function of the variable $r$ whose lower and the upper limits ${{r}_{1}}$ and ${{r}_{2}}$ are respectively written on the lower and the upper side of the sigma symbol. For expanding, the values of $r$ are substituted one by one into the function $f\left( r \right)$ and all the values are added together. The resulting series obtained is the required expanded form of the sum.
Complete step-by-step solution:
Let us consider a contracted form of a sum given by \[\sum\limits_{r={{r}_{1}}}^{{{r}_{2}}}{f\left( r \right)}\]. We can see that on the upper and the lower side of the sigma symbol, the values of the variable $r$ are present, ranging from ${{r}_{1}}$ to ${{r}_{2}}$. For expanding the sum, we have to put all the values of the variable $r$ from ${{r}_{1}}$ to ${{r}_{2}}$ one by one into the function $f\left( r \right)$ and all the obtained values from $f\left( {{r}_{1}} \right)$ to $f\left( {{r}_{2}} \right)$ are added together. The resulting series obtained is the required expanded form of the sum \[\sum\limits_{r={{r}_{1}}}^{{{r}_{2}}}{f\left( r \right)}\]. So we can write the expanded form of \[\sum\limits_{r={{r}_{1}}}^{{{r}_{2}}}{f\left( r \right)}\] as
$\Rightarrow \sum\limits_{r={{r}_{1}}}^{{{r}_{2}}}{f\left( r \right)}=f\left( {{r}_{1}} \right)+f\left( {{r}_{1}}+1 \right)+f\left( {{r}_{1}}+2 \right)+.......+f\left( {{r}_{2}}-1 \right)+f\left( {{r}_{2}} \right)$
Hence, this is the required way of writing a sum in the expanded form.
Note: It must be noted that the values of the limits of the variable of the summation must be all whole numbers. We can observe in the expansion shown above that the values of the variable are incremented by one, starting from the lower limit to the upper limit. This means that we have to cover only discrete natural numbers between ${{r}_{1}}$ to ${{r}_{2}}$, and not all of the real values. So the summation is discrete, not continuous. The case of continuous summation is the integration, in which all the real values between the lower and the upper limit are covered.
Complete step-by-step solution:
Let us consider a contracted form of a sum given by \[\sum\limits_{r={{r}_{1}}}^{{{r}_{2}}}{f\left( r \right)}\]. We can see that on the upper and the lower side of the sigma symbol, the values of the variable $r$ are present, ranging from ${{r}_{1}}$ to ${{r}_{2}}$. For expanding the sum, we have to put all the values of the variable $r$ from ${{r}_{1}}$ to ${{r}_{2}}$ one by one into the function $f\left( r \right)$ and all the obtained values from $f\left( {{r}_{1}} \right)$ to $f\left( {{r}_{2}} \right)$ are added together. The resulting series obtained is the required expanded form of the sum \[\sum\limits_{r={{r}_{1}}}^{{{r}_{2}}}{f\left( r \right)}\]. So we can write the expanded form of \[\sum\limits_{r={{r}_{1}}}^{{{r}_{2}}}{f\left( r \right)}\] as
$\Rightarrow \sum\limits_{r={{r}_{1}}}^{{{r}_{2}}}{f\left( r \right)}=f\left( {{r}_{1}} \right)+f\left( {{r}_{1}}+1 \right)+f\left( {{r}_{1}}+2 \right)+.......+f\left( {{r}_{2}}-1 \right)+f\left( {{r}_{2}} \right)$
Hence, this is the required way of writing a sum in the expanded form.
Note: It must be noted that the values of the limits of the variable of the summation must be all whole numbers. We can observe in the expansion shown above that the values of the variable are incremented by one, starting from the lower limit to the upper limit. This means that we have to cover only discrete natural numbers between ${{r}_{1}}$ to ${{r}_{2}}$, and not all of the real values. So the summation is discrete, not continuous. The case of continuous summation is the integration, in which all the real values between the lower and the upper limit are covered.
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