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EXERCISE 2.2 · Q35

Q.Find the sum to nn terms: 0.4+0.44+0.444+…0.4+0.44+0.444+\ldots

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tk=0.4+0.04+⋯+(k terms)=49(1−0.1k)t_k=0.4+0.04+\cdots+(k\text{ terms})=\dfrac49(1-0.1^k). $S_n=\sum_{k=1}^n t_k=\dfrac49\left[n-\sum_{k=1}^n0.1^k\right]=\dfrac49\left[n-\dfrac{1-0.1^n}9\right]=\dfrac{4n}9-\d …

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