If C0,C1,C2,.....Cn denotes the binomial coefficients in the expansion of (1+x)n, then C0+(C1/2)+(C2/3)+....+(Cn/n+1) is equal to

Question

If C0,C1,C2,.....Cn denotes the binomial coefficients in the expansion of (1+x)n, then C0+(C1/2)+(C2/3)+....+(Cn/n+1) is equal to (A)  (2n+1-1/n+1)  (B)  (2n-1/n)  (C)  (2n-1-1/n-1)  (D)  (2n+1-1/n+2)

Tardigrade Admin · Accepted Answer

We know, (1+x)n=C0+C1x+C2x2+....+Cnxn On integrating both sides 0 to 1, we get [ ((1+x)n+1/n+1) ]01 =[ C0x+(C1x2/2)+(C2x3/3)+....+(Cnxn+1/n+1) ]01 ⇒ (2n+1-1/n+1)=C0+(C1/2)+(C2/3)+.....+(Cn/n+1)

Q. If $C_{0}, C_{1}, C_{2}, ..... C_{n}$ denotes the binomial coefficients in the expansion of $(1 + x)^{n},$ then $C_{0} + \frac{C _{1}}{2} + \frac{C _{2}}{3} + .... + \frac{C _{n}}{n + 1}$ is equal to

$\frac{2 ^{n + 1} - 1}{n + 1}$

$\frac{2 ^{n} - 1}{n}$

$\frac{2 ^{n - 1} - 1}{n - 1}$

$\frac{2 ^{n + 1} - 1}{n + 2}$

Q. If C0​,C1​,C2​,.....Cn​ denotes the binomial coefficients in the expansion of (1+x)n, then C0​+2C1​​+3C2​​+....+n+1Cn​​ is equal to

n+12n+1−1​

n2n−1​

n−12n−1−1​

n+22n+1−1​

We know, (1+x)n=C0​+C1​x+C2​x2+....+Cn​xn On integrating both sides 0 to 1, we get [n+1(1+x)n+1​]01​ =[C0​x+2C1​x2​+3C2​x3​+....+n+1Cn​xn+1​]01​ ⇒ n+12n+1−1​=C0​+2C1​​+3C2​​+.....+n+1Cn​​

Q. If $C_{0}, C_{1}, C_{2}, ..... C_{n}$ denotes the binomial coefficients in the expansion of $(1 + x)^{n},$ then $C_{0} + \frac{C _{1}}{2} + \frac{C _{2}}{3} + .... + \frac{C _{n}}{n + 1}$ is equal to

$\frac{2 ^{n + 1} - 1}{n + 1}$

$\frac{2 ^{n} - 1}{n}$

$\frac{2 ^{n - 1} - 1}{n - 1}$

$\frac{2 ^{n + 1} - 1}{n + 2}$