Find the sum of 2n terms of a series of which every even term is a times the term before it, and every odd term c times the term before it, the first term being unity

(\frac{1 - a}{1 + a c}) (1 - a^{n} c^{n})

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(\frac{1 - a}{1 - a c}) (1 + a^{n} c^{n})

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(\frac{1 + a}{1 - a c}) (1 - a^{n} c^{n})

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(\frac{1 + a}{1 + a c}) (1 + a^{n} c^{n})

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Solution

The correct option is C $(\frac{1 + a}{1 - a c}) (1 - a^{n} c^{n})$
The series is $1 + a + a c + a^{2} c + a^{2} c^{2} + a^{3} c^{2} + . . . .$ to 2n terms
$\to (1 + a c + a^{2} c^{2} + \dots$ to n terms $)$ + $(a + a^{2} c + a^{3} c^{2} + \dots$ to n terms $)$
$= \frac{1 (1 - a^{n} c^{n})}{1 - a c} + \frac{a (1 - a^{n} c^{n})}{1 - a c} = (\frac{1 + a}{1 - a c}) (1 - a^{n} c^{n})$
Hence, option C.

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Find the sum of 2n terms of a series of which every even term is a times the term before it, and every odd term c times the term before it, the first term being unity

The correct option is C (1+a1−ac)(1−ancn)The series is 1+a+ac+a2c+a2c2+a3c2+.... to 2n terms →(1+ac+a2c2+… to n terms)+(a+a2c+a3c2+… to n terms)=1(1−ancn)1−ac+a(1−ancn)1−ac=(1+a1−ac)(1−ancn)Hence, option C.