If 1-x1-x21-x4 - 1-x128-r-0n xr- then n is equal to

(1+x)(1+x^2)(1+x^4) … (1+x^128)=∑_r=0^n x^r nbsp; Highest degree of x on both sides, n = 1+2+4+…… + 128 ⇒ n nbsp;=1+2+2^2 + …… +2^7 ⇒ n=2^8 12 1 ∴ nbsp; ...

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Standard XII

Mathematics

Sum of n Terms

If 1+x1+x21+x...

Question

If $(1 + x) (1 + x^{2}) (1 + x^{4}) \dots (1 + x^{128}) = n \sum r = 0 x^{r},$ then $n$ is equal to

256

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255

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254

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257

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Solution

The correct option is B $255$
$(1 + x) (1 + x^{2}) (1 + x^{4}) \dots (1 + x^{128}) = n \sum r = 0 x^{r}$
Highest degree of $x$ on both sides,
$n = 1 + 2 + 4 + \dots \dots + 128 \Rightarrow n = 1 + 2 + 2^{2} + \dots \dots + 2^{7} \Rightarrow n = \frac{2^{8} - 1}{2 - 1} ∴ n = 255$

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Sum of n Terms

Standard XII Mathematics

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If (1+x)(1+x2)(1+x4)…(1+x128)=n∑r=0xr, then n is equal to

The correct option is B 255(1+x)(1+x2)(1+x4)…(1+x128)=n∑r=0xr Highest degree of x on both sides, n=1+2+4+……+128⇒n=1+2+22+……+27⇒n=28−12−1∴n=255

If $(1 + x) (1 + x^{2}) (1 + x^{4}) \dots (1 + x^{128}) = n \sum r = 0 x^{r},$ then $n$ is equal to

The correct option is B $255$
$(1 + x) (1 + x^{2}) (1 + x^{4}) \dots (1 + x^{128}) = n \sum r = 0 x^{r}$
Highest degree of $x$ on both sides,
$n = 1 + 2 + 4 + \dots \dots + 128 \Rightarrow n = 1 + 2 + 2^{2} + \dots \dots + 2^{7} \Rightarrow n = \frac{2^{8} - 1}{2 - 1} ∴ n = 255$