Question

Evaluate the following $\sum\limits_{r = 1}^5 {{}^5{C_r}} $ .

Answer 1

Hint:
We have asked in the question to evaluate $\sum\limits_{r = 1}^5 {{}^5{C_r}} $ .
Since, we can write $\sum\limits_{r = 1}^5 {{}^5{C_r}} $ as $^5{C_1}{ + ^5}{C_2}{ + ^5}{C_3}{ + ^5}{C_4}{ + ^5}{C_5}$ .
Now, to solve further we will add and subtract ${}^5{C_0}$ in the above equation.
Then after, using property $\left( {{}^n{C_0} + {}^n{C_1} + ...................{ + ^n}{C_n}} \right) = {2^n}$ on the above equation and find the required answer.

Complete step by step solution:
We have asked in the question to evaluate $\sum\limits_{r = 1}^5 {{}^5{C_r}} $ .
We can write $\sum\limits_{r = 1}^5 {{}^5{C_r}} $ as $^5{C_1}{ + ^5}{C_2}{ + ^5}{C_3}{ + ^5}{C_4}{ + ^5}{C_5}$ .
 $\therefore \sum\limits_{r = 1}^5 {{}^5{C_r}} = {}^5{C_1}{ + ^5}{C_2}{ + ^5}{C_3}{ + ^5}{C_4}{ + ^5}{C_5}$ .
Now, to solve the above equation further, we will add and subtract ${}^5{C_0}$ in the above equation.
 $\therefore \sum\limits_{r = 1}^5 {{}^5{C_r}} = {}^5{C_0} + {}^5{C_1}{ + ^5}{C_2}{ + ^5}{C_3}{ + ^5}{C_4}{ + ^5}{C_5} - {}^5{C_0}$ .
Now, using property $\left( {{}^n{C_0} + {}^n{C_1} + ...................{ + ^n}{C_n}} \right) = {2^n}$ on the above equation, we get,
 $ = \left( {{}^5{C_0} + {}^5{C_1}{ + ^5}{C_2}{ + ^5}{C_3}{ + ^5}{C_4}{ + ^5}{C_5}} \right) - {}^5{C_0}$
$ = {2^5} - {}^5{C_0}$
Since, we know that the value of ${}^5{C_0} = 1$ .
$=32-1 \\
=31$

Hence, $\sum\limits_{r = 1}^5 {{}^5{C_r}} = 31$.

Note:
Sigma Notation: Sigma Notation is also known as summation notation and is a way to represent a sum of numbers. It is especially useful when the numbers have a specific pattern or would take too long to write out without abbreviation.
Some properties of sigma:
1) $\sum\limits_{n = h}^r {C.f\left( n \right) = C.} \sum\limits_{n = h}^r {f\left( n \right)} $ .
2) $\sum\limits_{n = h}^r {f\left( n \right) \pm \sum\limits_{n = h}^r {g\left( n \right) = } \sum\limits_{n = h}^r {\left( {f\left( n \right) \pm g\left( n \right)} \right)} } $ .
3) $\sum\limits_{i = 1}^n {C = nc} $ .
4) $\sum\limits_{i = 0}^n {i = \sum\limits_{i = 1}^n {i = } } \dfrac{{n\left( {n + 1} \right)}}{2}$ .
5) $\sum\limits_{i = 0}^n {{i^2} = } \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}$ .
 \[\sum\limits_{i = 0}^n {i = {{\left( {\sum\limits_{i = 0}^n i } \right)}^2}} = {\left( {\dfrac{{n\left( {n + 1} \right)}}{2}} \right)^2}\] .