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MISCELLANEOUS EXERCISE 4 (II) · Q94

Q.Prove, by method of induction, for all n∈Nn \in N: 8+17+26+…+(9n−1)=n2(9n+7)8 + 17 + 26 + \ldots + (9n-1) = \dfrac{n}{2}(9n+7).

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✓ Free question

Let P(n):8+17+⋯+(9n−1)=n2(9n+7)P(n):8+17+\cdots+(9n-1)=\dfrac n2(9n+7). Base: n=1n=1: L.H.S.=8=8, R.H.S.=12(16)=8=\dfrac12(16)=8; holds. Hypothesis: assume true for kk. Step: add (9(k+1)−1)=9k+8(9(k+1)-1)=9k+8: k2(9k+7)+9k+8=9k2+7k+18k+162=9k2+25k+162\dfrac k2(9k+7)+9k+8=\dfrac{9k^2+7k+18k+16}2=\dfrac{9k^2+25k+16}2. Target at k+1k+1: k+12(9k+16)=9k2+16k+9k+162=9k2+25k+162\dfrac{k+1}2(9k+16)=\dfrac{9k^2+16k+9k+16}2=\dfrac{9k^2+25k+16}2 — matches. Conclusion: true for all n∈Nn\in N.

✓Final answer

8+17+26+⋯+(9n−1)=n2(9n+7)8+17+26+\cdots+(9n-1)=\dfrac n2(9n+7), proved for all n∈Nn\in N.

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