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$If (1 + x) (1 + x^{2}) (1 + x^{4}) \dots (1 + x^{128}) = \sum_{i = 0}^{n} x^{r} then n is$

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Correct option is A

(1+x)1+x21+x4…1+x128=1+x+x2+…+xn1−x){(1+x)1+x21+x4…1+x128=(1−x)1+x+x2+..+xn→1−x256=1−xn+1→n+1=256n=255

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If (1+x)1+x21+x4…1+x128=∑i=0n xr then n is

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$If (1 + x) (1 + x^{2}) (1 + x^{4}) \dots (1 + x^{128}) = \sum_{i = 0}^{n} x^{r} then n is$