InterviewSolution
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InterviewSolution
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Find the sum of 2n terms of the series whose every even term is ‘a’ times the term before it and every odd term is ‘c’ times the term before it, the first term being unity. |
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Answer» Let T indicate a term of the progression.T1, T2, T3, ..., Tn, ...T2nT1 = 1 T2 = a T3 = ca T4 = c.a2 T5 = c2.a2 Tk if k is even = \(\frac{k}{a^2}.\frac{k}{C^2}-1\) T2n = \(\frac{2n}{a^2}\). \(\frac{2n}{C^2}-1\) T2n = an. Cn-1 S2n = 1 + a + ca + c.a2 + c2.a2 + c2.a3 .....an. cn - 1= 1 + [ a + c.a2 + c2.a3.... + an.cn - 1] + [ca + c2.a2 + c3.a3..... + cn - 1.an - 1]The sum of a G.P. = \(\frac{a(r^{n-1})}{r-1}\) For a + c.a2 + c2.a3.... + an.cn - 1 a = a, r = ca, n = n ⇒ \(\frac{a(ca^n-1)}{ca-1}\) For [ ca + c2.a2 + c3.a3..... + cn - 1.an - 1] a = ca, r = ca, n = n ⇒ \(\frac{ca(ca^n-1)}{ca-1}\) ∴ The required result = 1 + \(\frac{a(ca^n-1)}{ca-1}\) + \(\frac{ca(ca^n-1)}{ca-1}\) ⇒ \(\cfrac{a(ca^n-1)+ca(ca^{n-1}) + ca -1}{ca-1}\) |
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