Question

value of (50C0/1 + 50C2/3 + 50C4/5 + ... + 50C50/51) is:

A) (2^50)/51 B) (2^50 - 1)/51 C) (2^50 - 1)/50 D) (2^51 - 1)/51

Answer 1

$[\frac{ ^{50}C_0}{1} +\frac{ ^{50}C_2}{3}+\frac{^{50}C_4}{5}+......+\frac{ ^{50}C_{48 }}{49}+ \frac{^{50}C_{50} }{51}] = \frac{2^{50}}{51}$

Step-by-step explanation:

We know that

$(1+x)^{50}= ^{50}C_0 + ^{50}C_1x+ ^{50}C_2 x^{2} + ^{50}C_3 x^3+......+ ^{50}C_{49 }x^{49}+ ^{50}C_{50} x^{50}$

and $(1-x)^{50}= ^{50}C_0 - ^{50}C_1x+ ^{50}C_2 x^{2} - ^{50}C_3 x^3+......- ^{50}C_{49 }x^{49}+ ^{50}C_{50} x^{50}$

Therefore

$(1+x)^{50} +(1-x)^{50}$$= ^{50}C_0 + ^{50}C_1x+ ^{50}C_2 x^{2} + ^{50}C_3 x^3+......+ ^{50}C_{49 }x^{49}+ ^{50}C_{50} x^{50} +[(^{50}C_0 - ^{50}C_1x+ ^{50}C_2 x^{2} - ^{50}C_3 x^3+......- ^{50}C_{49 }x^{49}+ ^{50}C_{50} x^{50}]$

⇔$(1+x)^{50} +(1-x)^{50}$$= 2[^{50}C_0+ ^{50}C_ x^2+^{50}C_4x^4+......+ ^{50}C_{48 }x^{48}+ ^{50}C_{50} x^{50}]$

Integrating both sides

$\int^1_0[(1+x)^{50} +(1-x)^{50}]dx$$=2\int ^1_0[ ^{50}C_0+ ^{50}C_ x^2+^{50}C_4x^4+......+ ^{50}C_{48 }x^{48}+ ^{50}C_{50} x^{50}]dx$

⇔$[\frac{(1+x)^{51}}{51}]^1_0- [\frac{(1-x)^{51}}{51}]^1_0$$=2[ ^{50}C_0\frac{x}{1} + ^{50}C_2\frac{ x^3}{3}+^{50}C_4\frac{x^5}{5}+......+ ^{50}C_{48 }\frac{x^{49}}{49}+ ^{50}C_{50} \frac{x^{51}}{51}]^1_0$

⇔$\frac{2^{51}}{51} -\frac{1}{51}+\frac{1}{51}$$=2[\frac{ ^{50}C_0}{1} +\frac{ ^{50}C_2}{3}+\frac{^{50}C_4}{5}+......+\frac{ ^{50}C_{48 }}{49}+ \frac{^{50}C_{50} }{51}]$

⇔$\frac{2^{51}}{51} =2[\frac{ ^{50}C_0}{1} +\frac{ ^{50}C_2}{3}+\frac{^{50}C_4}{5}+......+\frac{ ^{50}C_{48 }}{49}+ \frac{^{50}C_{50} }{51}]$

⇔$[\frac{ ^{50}C_0}{1} +\frac{ ^{50}C_2}{3}+\frac{^{50}C_4}{5}+......+\frac{ ^{50}C_{48 }}{49}+ \frac{^{50}C_{50} }{51}] = \frac{2^{50}}{51}$