MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Expansions to Remove Indeterminate Form

If 1+x1+x21...

Question

If $(1 + x) (1 + x^{2}) (1 + x^{4}) . . . . (1 + x^{128}) = \sum_{r = 0}^{n} x^{r}$ then findthe value of $n$ .

Open in App

Solution

$(1 + x) (1 + x^{2}) (1 + x^{4}) . . . . . (1 + x^{128}) = n \sum r = 0 x^{r}$
$(1 + x^{2 n})$ so $n = \frac{128}{2} = 64$

Suggest Corrections

thumbs-up

0

Similar questions

(1 + x) (1 + x^{2}) (1 + x^{4}) . . . . . (1 + x^{128}) = n \sum r = 0 x^{r}

then find the value of

n

(1 + x) (1 + x^{2}) (1 + x^{4}) . . . (1 + x^{128}) = \sum_{r = 0}^{n} x^{r}

n

(1 + x) (1 + x^{2}) (1 + x^{4}) \dots (1 + x^{128}) = n \sum r = 0 x^{r},

n

(1 + x) (1 + x^{2}) (1 + x^{4}) \dots (1 + x^{128}) = n \sum r = 0 x^{r},

n

(1 + x) (1 + x^{2}) (1 + x^{4}) . . . .

(1 + x^{128}) = (1 + x + x^{2} . . . . . x^{n})

n = ?

| x | < 1

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Standard Expansions and Standard Limits

MATHEMATICS

Watch in App

explore_more_icon

Explore more

Expansions to Remove Indeterminate Form

Standard XII Mathematics

Join BYJU'S Learning Program

CrossIcon