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Finite sum of terms a^k b^(n-k) using a G.P.

This type of question, asked 3 times: 25 Jun 2022, 28 Jul 2022, 28 Jan 2026 S2. All 3 questions below, oldest first.

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  1. Question 1

    25 Jun 2022

    68
    If 12⋅310+122⋅39+…1210⋅3=K210⋅310\frac{1}{2 \cdot3^{10}}+\frac{1}{2^{2}\cdot3^{9}}+ \ldots\frac{1}{2^{10}\cdot 3}=\frac{K}{2^{10}\cdot3^{10}}, then the remainder when KK is divided by 6 is

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  2. Question 2

    28 Jul 2022

    176
    If 6312+10311+20310+4039+…..+102403=2n⋅m\frac{6}{3^{12}}+\frac{10}{3^{11}}+\frac{20}{3^{10}}+\frac{40}{3^{9}}+ \ldots.. +\frac{10240}{3}=2^{n}\cdot m, where mm is odd, then m.nm.n is equal to

  3. Question 3

    28 Jan 2026 S2

    70
    6326+10.1325+10.2324+10.22323+…+10.2243\frac{6}{3^{26}}+\frac{10.1}{3^{25}}+\frac{10.2}{3^{24}}+\frac{10.2^{2}}{3^{23}}+ \ldots +\frac{10.2^{24}}{3} is equal to

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