$1 + r + r^{2} + r^{3} + . . . . . . . + r^{n} = (1 + r) (1 + r^{2}) (1 + r^{4}) (1 + r^{8})$ , then the value of n is :

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Solution

The correct option is C 15
Given: $1 + r + r^{2} + r^{3} + . . . + r^{n} = (1 + r) (1 + r^{2}) (1 + r^{4}) (1 + r^{8})$

Use summation formulae for G.P series $1 + r + r^{2} + r^{3} + . . . + r^{n}$ which is $(S_{n} = a (\frac{r^{n} - 1}{r - 1}))$

$1 + r + r^{2} + r^{3} + . . . + r^{n} = S_{n} = 1 (\frac{r^{n} - 1}{r - 1}) . . . . . . E q : 01$

Simplify the series: $(1 + r) (1 + r^{2}) (1 + r^{4}) (1 + r^{8})$

${(1 + r) (1 + r^{2})} {(1 + r^{4}) (1 + r^{8})}$

$(1 + r^{2} + r + r^{3}) (1 + r^{8} + r^{4} + r^{12})$

$1 + r^{8} + r^{4} + r^{12} + r^{2} + r^{10} + r^{6} + r^{14} + r + r^{9} + r^{5} + r^{13} + r^{3} + r^{11} + r^{7} + r^{15}$

$1 + r + r^{2} + . . . + r^{14} + r^{15}$

Since, it is a geometric progression where the number of terms is 15. Use the summation formula to find
the summation up to 15 terms.
$S_{15} = 1 (\frac{r^{15} - 1}{r - 1})$ ......Eq:02
Compare Eq:01 and Eq:02;

$1 (\frac{r^{n} - 1}{r - 1})$

$1 (\frac{r^{15} - 1}{r - 1})$

$n = 15$

Option (C) is correct.

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Similar questions

1 + r + r^{2} + \dots . . . + r^{n} =

(1 + r) (1 + r^{2}) (1 + r^{4}) (1 + r^{8}),

n

(\frac{1}{r_{1}} + \frac{1}{r_{2}}) (\frac{1}{r_{2}} + \frac{1}{r_{3}}) (\frac{1}{r_{3}} + \frac{1}{r_{1}}) = \frac{64 R^{3}}{a^{2} b^{2} c^{2}}

Δ A B C (\frac{1}{r_{1}} + \frac{1}{r_{2}}) (\frac{1}{r_{2}} + \frac{1}{r_{3}}) (\frac{1}{r_{3}} + \frac{1}{r_{1}}) = \frac{K R^{3}}{a^{2} b^{2} c^{2}}

K

△ A B C,

(\frac{1}{r_{1}} + \frac{1}{r_{2}}) (\frac{1}{r_{2}} + \frac{1}{r_{3}}) (\frac{1}{r_{3}} + \frac{1}{r_{1}}) = \frac{K R^{3}}{a^{2} b^{2} c^{2}},

K

r_{1}, r_{2}, r_{3}

\frac{1}{r_{1}} + \frac{1}{r_{2}} + \frac{1}{r_{3}} - \frac{1}{r}

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