The sum of the series 20C0-20C1+20C2-20C3+ ldots +20C10 is equal to

Question

The sum of the series 20C0-20C1+20C2-20C3+ ldots +20C10 is equal to (A) -20C10 (B) (1/2) 20C10 (C) 0 (D) 20C10

Tardigrade Admin · Accepted Answer

S= 20 C0- 20 C1+ 20 C2+ ldots+ 20 C10 We know that, 20 C0- 20 C1+ 20 C2- ldots+ 20 C20=0 ⇒ 220 C0- 20 C1+ 20 C2+ ldots- 20 C9+ 20 C10=0 ⇒ 20 C0- 20 C1+ 20 C2 ldots- 20 C9=-(1/2) 20 C10 So, S=-(1/2) 20 C10+ 20 C10=(1/2) 20 C10

Q. The sum of the series $^{20} C_{0} -^{20} C_{1} +^{20} C_{2} -^{20} C_{3} + \dots +^{20} C_{10}$ is equal to

$-^{20} C_{10}$

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

$0$

$^{20} C_{10}$

Q. The sum of the series 20C0​−20C1​+20C2​−20C3​+…+20C10​ is equal to

−20C10​

21​20C10​

0

20C10​

S=20C0​−20C1​+20C2​+…+20C10​ We know that, 20C0​−20C1​+20C2​−…+20C20​=0 ⇒220C0​−20C1​+20C2​+…−20C9​+20C10​=0 ⇒20C0​−20C1​+20C2​…−20C9​=−21​20C10​ So, S=−21​20C10​+20C10​=21​20C10​

Q. The sum of the series $^{20} C_{0} -^{20} C_{1} +^{20} C_{2} -^{20} C_{3} + \dots +^{20} C_{10}$ is equal to

$-^{20} C_{10}$

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

$0$

$^{20} C_{10}$