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Q.Find the sum to nn terms of the sequence 4+44+444+…4 + 44 + 444 + \ldots

Kerala DhseKerala DHSE Plus One Board 2018Subjective· 3mImportance★★★★★
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Each term is a repunit-type number; write the k-th term as 4·(10^k − 1)/9 and sum the geometric part to get S_n = (4/81)(10^(n+1) − 9n − 10).

The series 4 + 44 + 444 + … is NOT itself a GP, so we rewrite each term using repunits.

Take out 4: S_n = 4(1 + 11 + 111 + … to n terms).

The k-th repunit 1, 11, 111, … equals (10^k − 1)/9, because 111…1 (k ones) = (10^k − 1)/9.

So S_n = (4/9) · Σ (from k=1 to n) (10^k − 1) = (4/9)[ (10 + 10^2 + … + 10^n) − n ].

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