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$\sum_{r = 0}^{n} (- 1)^{r} (^{n} C_{r}) \frac{1 + rx}{1 + nx}$ is equal to

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Correct option is D

Let E=∑r=0n (−1)r nCr1+rx1+nx=11+nx∑r=0n (−1)r nCr(1+rx)=11+nx∑r=0n (−1)r⋅nCr+x∑r=0n r(−1)rnCr=11+nx(0+0)=0∵nC0−nC1+nC2−nC3+…(−1)n nCn=0

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∑r=0n (−1)r nCr1+rx1+nx is equal to

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$\sum_{r = 0}^{n} (- 1)^{r} (^{n} C_{r}) \frac{1 + rx}{1 + nx}$ is equal to