Question

(1 + x) (1 + x²) (1 + x⁴) .........(1 + x¹²⁸) = $\sum_{r = 0}^{n}x^{r}$ then $\frac{n}{85}$=

• A. 2
• B. 3
• C. 1
• D. 9

Answer 1

Answer: (B)

3

Answer 2

(1 + x) (1 + x²) (1+x⁴) ....(1+x¹²⁸) = $\frac{1(1–x^{n + 1})}{1–x}$

(1 – xⁿ⁺¹) = (1 – x)[(1 + x) (1 + x²) (1 + x⁴)....(1+ x¹²⁸)]1 – xⁿ⁺¹

= 1 – x²⁵⁶

Ž n + 1 = 256 Ž n = 255 Ž $\frac{n}{85}$ = $\frac{255}{85}$ = 3