If 1 - r - r 2 - - - - - - r n - 1 - r 1 - r 2 1 -r 4 1 - r 8 - then the value of n is

The correct options are A 13 D 16 1+r+r^2+........+r^n=(1+r)(1+r^2)(1+r^4)(1+r^8) or>(1 r^n1 r)=(1+r)(1+r^2)(1+r^4)(1+r^8) or> 1 ...

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Sum of Powers

If 1 + r + ...

Question

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

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1 + r + r^{2} + r^{3} + . . . . . . . + r^{n} = (1 + r) (1 + r^{2}) (1 + r^{4}) (1 + r^{8})

△ 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

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

(\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}}

R_{1}

R_{2}

\frac{n}{1}

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