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Question: Evaluate the following \(\sum\limits_{r = 1}^5 {{}^5{C_r}} \) ....

Evaluate the following ∑r=155Cr\sum\limits_{r = 1}^5 {{}^5{C_r}} .

Explanation

Solution

We have asked in the question to evaluate ∑r=155Cr\sum\limits_{r = 1}^5 {{}^5{C_r}} .
Since, we can write ∑r=155Cr\sum\limits_{r = 1}^5 {{}^5{C_r}} as 5C1+5C2+5C3+5C4+5C5^5{C_1}{ + ^5}{C_2}{ + ^5}{C_3}{ + ^5}{C_4}{ + ^5}{C_5} .
Now, to solve further we will add and subtract 5C0{}^5{C_0} in the above equation.
Then after, using property (nC0+nC1+...................+nCn)=2n\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 ∑r=155Cr\sum\limits_{r = 1}^5 {{}^5{C_r}} .
We can write ∑r=155Cr\sum\limits_{r = 1}^5 {{}^5{C_r}} as 5C1+5C2+5C3+5C4+5C5^5{C_1}{ + ^5}{C_2}{ + ^5}{C_3}{ + ^5}{C_4}{ + ^5}{C_5} .
∴∑r=155Cr=5C1+5C2+5C3+5C4+5C5\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 5C0{}^5{C_0} in the above equation.
∴∑r=155Cr=5C0+5C1+5C2+5C3+5C4+5C5−5C0\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 (nC0+nC1+...................+nCn)=2n\left( {{}^n{C_0} + {}^n{C_1} + ...................{ + ^n}{C_n}} \right) = {2^n} on the above equation, we get,
=(5C0+5C1+5C2+5C3+5C4+5C5)−5C0= \left( {{}^5{C_0} + {}^5{C_1}{ + ^5}{C_2}{ + ^5}{C_3}{ + ^5}{C_4}{ + ^5}{C_5}} \right) - {}^5{C_0}
=25−5C0= {2^5} - {}^5{C_0}
Since, we know that the value of 5C0=1{}^5{C_0} = 1 .
=32−1 =31=32-1 \\\ =31

Hence, ∑r=155Cr=31\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. ∑n=hrC.f(n)=C.∑n=hrf(n)\sum\limits_{n = h}^r {C.f\left( n \right) = C.} \sum\limits_{n = h}^r {f\left( n \right)} .
  2. ∑n=hrf(n)±∑n=hrg(n)=∑n=hr(f(n)±g(n))\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. ∑i=1nC=nc\sum\limits_{i = 1}^n {C = nc} .
  4. ∑i=0ni=∑i=1ni=n(n+1)2\sum\limits_{i = 0}^n {i = \sum\limits_{i = 1}^n {i = } } \dfrac{{n\left( {n + 1} \right)}}{2} .
  5. ∑i=0ni2=n(n+1)(2n+1)6\sum\limits_{i = 0}^n {{i^2} = } \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6} .
    ∑i=0ni=(∑i=0ni)2=(n(n+1)2)2\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} .