The sum ${}^{20}{C_0} + {}^{20}{C_1} + {}^{20}{C_2} + .....{}^{20}{C_{10}}$ is equal to
A.${2^{20}} + \dfrac{{20!}}{{2{{\left( {10!} \right)}^2}}}$
B.${2^{19}} + \dfrac{1}{2}.\dfrac{{20!}}{{{{\left( {10!} \right)}^2}}}$
C.${2^{19}} + {}^{20}{C_{10}}$
D.None of these
Answer
653.7k+ views
Hint: The given terms are coefficients of binomial expansion of (1+x)n , where n=20. Taking x=1 , we get the sum of the coefficient. Using the relation ${}^n{C_r} = {}^n{C_{n - r}}$ and further simplification, we get the required solution.
Complete step-by-step answer:
We know that binomial expansion of a + b to the power n is given by \[{\left( {a{\text{ }} + {\text{ }}b} \right)^n} = {\text{ }}\left( {^n{C_0}} \right){a^n}{b^0} + {\text{ }}\left( {^n{C_1}} \right){a^{n - 1}}b{\text{ }} + {\text{ }}\left( {^n{C_2}} \right){a^{n - 2}}{b^2} + {\text{ }} \ldots {\text{ }} + {\text{ }}\left( {^n{C_{n - 1}}} \right)a{b^{n - 1}} + {\text{ }}\left( {^n{C_n}} \right){a^0}{b^n}\]
Using the above equation of binomial expansion, the expansion of ${\left( {1 + x} \right)^{20}}$, n=20,is given by,
\[{\left( {1 + x} \right)^{20}} = {\text{ }}\left( {^{20}{C_0}} \right){x^0} + {\text{ }}\left( {^{20}{C_1}} \right){x^1}{\text{ }} + {\text{ }}\left( {^{20}{C_2}} \right){x^2} + {\text{ }} \ldots {\text{ }} + {\text{ }}\left( {^{20}{C_{19}}} \right){x^{19}} + {\text{ }}\left( {^{20}{C_{20}}} \right){x^{20}}\]
Taking x= 1, the above binomial expansion becomes,
\[{\left( {1 + 1} \right)^{20}} = {\text{ }}\left( {^{20}{C_0}} \right) + {\text{ }}\left( {^{20}{C_1}} \right){\text{ }} + {\text{ }}\left( {^{20}{C_2}} \right) + {\text{ }} \ldots {\text{ }} + {\text{ }}\left( {^{20}{C_{19}}} \right) + {\text{ }}\left( {^{20}{C_{20}}} \right)\]
Now we have the sum of coefficients as the 20th power of 2.
We know that, ${}^n{C_r} = {}^n{C_{n - r}}$, using this in above equation, we get,
\[{\left( 2 \right)^{20}} = {\text{ 2}} \times \left[ {\left( {^{20}{C_0}} \right) + {\text{ }}\left( {^{20}{C_1}} \right){\text{ }} + {\text{ }}\left( {^{20}{C_2}} \right) + {\text{ }} \ldots {\text{ }} + {\text{ }}\left( {^{20}{C_9}} \right) + \left( {^{20}{C_{10}}} \right)} \right] - \left( {^{20}{C_{10}}} \right)\]
On adding both sides with \[\left( {^{20}{C_{10}}} \right)\], we get
\[{\left( 2 \right)^{20}} + \left( {^{20}{C_{10}}} \right) = {\text{ 2}} \times \left[ {\left( {^{20}{C_0}} \right) + {\text{ }}\left( {^{20}{C_1}} \right){\text{ }} + {\text{ }}\left( {^{20}{C_2}} \right) + {\text{ }} \ldots {\text{ }} + {\text{ }}\left( {^{20}{C_9}} \right) + \left( {^{20}{C_{10}}} \right)} \right]\]
Dividing both sides with 2 and reversing the equation, we get
\[\left[ {\left( {^{{\text{20}}}{{\text{C}}_{\text{0}}}} \right){\text{ + }}\left( {^{{\text{20}}}{{\text{C}}_{\text{1}}}} \right){\text{ + }}\left( {^{{\text{20}}}{{\text{C}}_{\text{2}}}} \right){\text{ }} \ldots {\text{ }}+{\text{ }}\left( {^{{\text{20}}}{{\text{C}}_{\text{9}}}} \right) + \left( {^{20}{C_{10}}} \right)} \right]{\text{ = }}\dfrac{{\text{1}}}{{\text{2}}}\left( {{{\left( {\text{2}} \right)}^{{\text{20}}}} + \left( {^{{\text{20}}}{{\text{C}}_{{\text{10}}}}} \right)} \right)\]
By opening the bracket and further simplification, we get,
${\text{ = }}{{\text{2}}^{{\text{19}}}} + \dfrac{{\text{1}}}{{\text{2}}}{\text{.}}\left( {^{{\text{20}}}{{\text{C}}_{{\text{10}}}}} \right)$
Therefore, the sum of given terms of binomial expansions is given by,
${\text{ = }}{{\text{2}}^{{\text{19}}}}{\text{ + }}\dfrac{{\text{1}}}{{\text{2}}}{\text{.}}\dfrac{{{\text{20!}}}}{{{{\left( {{\text{10!}}} \right)}^{\text{2}}}}}$
This answer is not in any of the options. So, we can mark none of these.
So the correct answer is option D
Note: The coefficients of binomial expansion of power n is the nth row of the pascal's triangle. The solution can also be started from taking the binomial expansion of 1 + 1 to the power n (20 in this case). Concept of permutations and combinations are essential for binomial expansion, especially to find out the coefficients. But these concepts are not used in this particular problem. A common error while doing this problem is that the middle term in the expansion \[\left( {^{20}{C_{10}}} \right)\] is not included in the bracket after using the relation ${}^n{C_r} = {}^n{C_{n - r}}$. It must be included in the bracket by changing the equation appropriately.
Complete step-by-step answer:
We know that binomial expansion of a + b to the power n is given by \[{\left( {a{\text{ }} + {\text{ }}b} \right)^n} = {\text{ }}\left( {^n{C_0}} \right){a^n}{b^0} + {\text{ }}\left( {^n{C_1}} \right){a^{n - 1}}b{\text{ }} + {\text{ }}\left( {^n{C_2}} \right){a^{n - 2}}{b^2} + {\text{ }} \ldots {\text{ }} + {\text{ }}\left( {^n{C_{n - 1}}} \right)a{b^{n - 1}} + {\text{ }}\left( {^n{C_n}} \right){a^0}{b^n}\]
Using the above equation of binomial expansion, the expansion of ${\left( {1 + x} \right)^{20}}$, n=20,is given by,
\[{\left( {1 + x} \right)^{20}} = {\text{ }}\left( {^{20}{C_0}} \right){x^0} + {\text{ }}\left( {^{20}{C_1}} \right){x^1}{\text{ }} + {\text{ }}\left( {^{20}{C_2}} \right){x^2} + {\text{ }} \ldots {\text{ }} + {\text{ }}\left( {^{20}{C_{19}}} \right){x^{19}} + {\text{ }}\left( {^{20}{C_{20}}} \right){x^{20}}\]
Taking x= 1, the above binomial expansion becomes,
\[{\left( {1 + 1} \right)^{20}} = {\text{ }}\left( {^{20}{C_0}} \right) + {\text{ }}\left( {^{20}{C_1}} \right){\text{ }} + {\text{ }}\left( {^{20}{C_2}} \right) + {\text{ }} \ldots {\text{ }} + {\text{ }}\left( {^{20}{C_{19}}} \right) + {\text{ }}\left( {^{20}{C_{20}}} \right)\]
Now we have the sum of coefficients as the 20th power of 2.
We know that, ${}^n{C_r} = {}^n{C_{n - r}}$, using this in above equation, we get,
\[{\left( 2 \right)^{20}} = {\text{ 2}} \times \left[ {\left( {^{20}{C_0}} \right) + {\text{ }}\left( {^{20}{C_1}} \right){\text{ }} + {\text{ }}\left( {^{20}{C_2}} \right) + {\text{ }} \ldots {\text{ }} + {\text{ }}\left( {^{20}{C_9}} \right) + \left( {^{20}{C_{10}}} \right)} \right] - \left( {^{20}{C_{10}}} \right)\]
On adding both sides with \[\left( {^{20}{C_{10}}} \right)\], we get
\[{\left( 2 \right)^{20}} + \left( {^{20}{C_{10}}} \right) = {\text{ 2}} \times \left[ {\left( {^{20}{C_0}} \right) + {\text{ }}\left( {^{20}{C_1}} \right){\text{ }} + {\text{ }}\left( {^{20}{C_2}} \right) + {\text{ }} \ldots {\text{ }} + {\text{ }}\left( {^{20}{C_9}} \right) + \left( {^{20}{C_{10}}} \right)} \right]\]
Dividing both sides with 2 and reversing the equation, we get
\[\left[ {\left( {^{{\text{20}}}{{\text{C}}_{\text{0}}}} \right){\text{ + }}\left( {^{{\text{20}}}{{\text{C}}_{\text{1}}}} \right){\text{ + }}\left( {^{{\text{20}}}{{\text{C}}_{\text{2}}}} \right){\text{ }} \ldots {\text{ }}+{\text{ }}\left( {^{{\text{20}}}{{\text{C}}_{\text{9}}}} \right) + \left( {^{20}{C_{10}}} \right)} \right]{\text{ = }}\dfrac{{\text{1}}}{{\text{2}}}\left( {{{\left( {\text{2}} \right)}^{{\text{20}}}} + \left( {^{{\text{20}}}{{\text{C}}_{{\text{10}}}}} \right)} \right)\]
By opening the bracket and further simplification, we get,
${\text{ = }}{{\text{2}}^{{\text{19}}}} + \dfrac{{\text{1}}}{{\text{2}}}{\text{.}}\left( {^{{\text{20}}}{{\text{C}}_{{\text{10}}}}} \right)$
Therefore, the sum of given terms of binomial expansions is given by,
${\text{ = }}{{\text{2}}^{{\text{19}}}}{\text{ + }}\dfrac{{\text{1}}}{{\text{2}}}{\text{.}}\dfrac{{{\text{20!}}}}{{{{\left( {{\text{10!}}} \right)}^{\text{2}}}}}$
This answer is not in any of the options. So, we can mark none of these.
So the correct answer is option D
Note: The coefficients of binomial expansion of power n is the nth row of the pascal's triangle. The solution can also be started from taking the binomial expansion of 1 + 1 to the power n (20 in this case). Concept of permutations and combinations are essential for binomial expansion, especially to find out the coefficients. But these concepts are not used in this particular problem. A common error while doing this problem is that the middle term in the expansion \[\left( {^{20}{C_{10}}} \right)\] is not included in the bracket after using the relation ${}^n{C_r} = {}^n{C_{n - r}}$. It must be included in the bracket by changing the equation appropriately.
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