Question 274 of 347MediumMCQ
The sum of the series 20C0−20C1+20C2−20C3+⋯−⋯+20C10 is:
Solutions · 1
Official solution
We know
\[{}^{20}{C_0} - {}^{20}{C_1} + {}^{20}{C_2} - {}^{20}{C_3} + .... + {}^{20}{C_{10}} - {}^{20}{C_{11}}+ ...... + {}^{20}{C_{20}} = 0\]
\[ \Rightarrow \] \[({}^{20}{C_0} - {}^{20}{C_1} + {}^{20}{C_2} - {}^{20}{C_3} + .... - {}^{20} {C_{9}})\] +20C10 \[(-{}^{20}{C_{9}}+ {}^{20}{C_{8}}+...... + {}^{20}{C_{0}})\]
(As 20C11=20C9)
\[ \Rightarrow \] \[2({}^{20}{C_0} - {}^{20}{C_1} + {}^{20}{C_2} - {}^{20}{C_3} + .... - {}^{20} {C_{9}})\] +20C10 = 0
\[ \Rightarrow \] \[{}^{20}{C_0} - {}^{20}{C_1} + {}^{20}{C_2} - {}^{20}{C_3} + .... - {}^{20} {C_{9}}\] = \[ - {1 \over 2}{}^{20}{C_{10}}\]
Adding 20C10 both sides,
\[ \Rightarrow \] \[{}^{20}{C_0} - {}^{20}{C_1} + {}^{20}{C_2} - {}^{20}{C_3} + .... - {}^{20} {C_{9}}\] +20C10 = \[ - {1 \over 2}{}^{20}{C_{10}}\] +20C10
\[ \Rightarrow \] \[{}^{20}{C_0} - {}^{20}{C_1} + {}^{20}{C_2} - {}^{20}{C_3} + .... - {}^{20} {C_{9}}\] +20C10 = \[ {1 \over 2}{}^{20}{C_{10}}\]
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