The value of $^{20} C_{0} +^{20} C_{1} +^{20} C_{2} +^{20} C_{3} +^{20} C_{4} +^{20} C_{12} +^{20} C_{13} +^{20} C_{14} +^{20} C_{15}$ is

2^{19} - \frac{(^{20} C_{10} +^{20} C_{9})}{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2^{19} - \frac{(^{20} C_{10} + 2 \times^{20} C_{9})}{2}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

2^{19} - \frac{^{20} C_{10}}{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

none of these

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $2^{19} - \frac{(^{20} C_{10} + 2 \times^{20} C_{9})}{2}$
Consider $(1 + x)^{20}$

$(1 + x)^{20} =^{20} C_{0} +^{20} C_{1} x +^{20} C_{2} x^{2} . . .^{20} C_{20} x^{20}$

Substituting $x = 1$ we get

$2^{20} =^{20} C_{0} +^{20} C_{1} +^{20} C_{2} . . .^{20} C_{20}$

By using the property of the binomial coefficient

$^{n} C_{r} =^{n} C_{n - r}$ we get

$2^{20} = 2 (^{20} C_{0} +^{20} C_{1} +^{20} C_{2} +^{20} C_{3} +^{20} C_{4} +^{20} C_{5} +^{20} C_{6} +^{20} C_{7} +^{20} C_{8} +^{20} C_{9}) +^{20} C_{10}$

$^{20} C_{10}$ is left out since it is the middle term.

Hence

$2^{20} -^{20} C_{10} = 2 (^{20} C_{0} +^{20} C_{1} +^{20} C_{2} +^{20} C_{3} +^{20} C_{4} +^{20} C_{5} +^{20} C_{6} +^{20} C_{7} +^{20} C_{8} +^{20} C_{9})$

$2^{19} - \frac{^{20} C_{10}}{2} =^{20} C_{0} +^{20} C_{1} +^{20} C_{2} +^{20} C_{3} +^{20} C_{4} +^{20} C_{5} +^{20} C_{6} +^{20} C_{7} +^{20} C_{8} +^{20} C_{9}$

$2^{19} - \frac{^{20} C_{10} + 2^{20} C_{9}}{2} =^{20} C_{0} +^{20} C_{1} +^{20} C_{2} +^{20} C_{3} +^{20} C_{4} +^{20} C_{5} +^{20} C_{6} +^{20} C_{7} +^{20} C_{8}$

Again applying $^{n} C_{r} =^{n} C_{n - r}$ we get

$2^{19} - \frac{^{20} C_{10} + 2^{20} C_{9}}{2} =^{20} C_{0} +^{20} C_{1} +^{20} C_{2} +^{20} C_{3} +^{20} C_{4} +^{20} C_{15} +^{20} C_{14} +^{20} C_{13} +^{20} C_{12}$

Suggest Corrections