The sum of the series ^20C_0 - ^20C_1 + ^20C_2 - ^20C_3 + + ^20C_10 is: - Mathematics MCQ AIEEE 2007 2026 | ExamTest.live

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The sum of the series^{20} C_{0} -^{20} C_{1} +^{20} C_{2} -^{20} C_{3} + \dots +^{20} C_{10} is:

Solution & Step-by-step Explanation

Consider the expansion: (1 - 1)^{20} =^{20} C_{0} -^{20} C_{1} +^{20} C_{2} - \dots +^{20} C_{20} = 0 .Using the property^{n} C_{r} =^{n} C_{n - r} :^{20} C_{0} =^{20} C_{20},^{20} C_{1} =^{20} C_{19}, \dots,^{20} C_{9} =^{20} C_{11} The total sum is: (^{20} C_{0} -^{20} C_{1} + \dots -^{20} C_{9}) +^{20} C_{10} - (^{20} C_{11} - \dots +^{20} C_{20}) = 0 Since terms are symmetric: 2 (^{20} C_{0} -^{20} C_{1} + \dots -^{20} C_{9}) +^{20} C_{10} = 0^{20} C_{0} -^{20} C_{1} + \dots -^{20} C_{9} = - \frac{1}{2}^{20} C_{10} The required sum is S = (^{20} C_{0} -^{20} C_{1} + \dots -^{20} C_{9}) +^{20} C_{10} $ S = - \frac{1}{2}^{20} C_{10} +^{20} C_{10} = \frac{1}{2}^{20} C_{10} $

Try it yourself before checking the explanation above.

The sum of the series^{20} C_{0} -^{20} C_{1} +^{20} C_{2} -^{20} C_{3} + \dots +^{20} C_{10} is:

-^{20} C_{10}

\frac{1}{2}^{20} C_{10}

0

^{20} C_{10}

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