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If $\sum_{k = 1}^{n} ϕ (k) = \frac{2 n}{n + 1},$ then $\sum_{k = 1}^{10} \frac{1}{ϕ (k)}$ is equal to

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Correct option is B

Let Sn=∑k=1n ϕ(k)=2nn+1⇒ϕ(n)=Sn−Sn−1=2nn+1−2(n−1)n=2n2−n2−1(n+1)n=2n(n+1)⇒1ϕ(n)=n(n+1)2⇒∑k=110 1ϕ(n)=12∑k=110 k2+k=12⋅16(10)(11)(21)+12⋅12(10)(11)=220

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If ∑k=1n ϕ(k)=2nn+1, then ∑k=110 1ϕ(k) is equal to

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If $\sum_{k = 1}^{n} ϕ (k) = \frac{2 n}{n + 1},$ then $\sum_{k = 1}^{10} \frac{1}{ϕ (k)}$ is equal to