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MISCELLANEOUS EXERCISE - 2 (II) · Q111

Q.Find 2+22+222+2222+⋯2+22+222+2222+\cdots upto nn terms.

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2+22+222+⋯=2(1+11+111+⋯ )=2×10n+1−10−9n812+22+222+\cdots=2(1+11+111+\cdots)=2\times\dfrac{10^{n+1}-10-9n}{81} (reusing the repunit-sum result). …

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