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21 February, 13:11

Use mathematical induction to prove the statement is true for all positive integers n.

10 + 30 + 60 + ... + 10n = 5n (n + 1)

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  1. 21 February, 14:30
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    F (n) = 10 + 30 + 60 + ... + 10n = 5n (n + 1)

    1. Let n = 1.

    LHS = 10n = 10 * 1 = 10 [ LHS - Left hand side ]

    RHS = 5n (n + 1) = 5*1*2 = 10 [ RHS - Right hand side ]

    LHS = RHS

    Hence, f (n) is valid for n = 1.

    2. Supposing that f (n) is valid for m.

    So,

    10 + 30 + 60 + ... + 10m = 5m (m + 1)

    3. Let n = m + 1

    LHS = 10 + 30 + 60 + ... + 10 (m + 1) = 10 + 30 + 60 + ... + 10m + 10

    RHS = 5 (m + 1) (m + 1 + 1) = 5 (m + 1) (m + 2) = 5m (m + 1) + 10

    We know from step 2, that 10 + 30 + 60 + ... + 10m = 5m (m + 1).

    Hence,

    10 + 30 + 60 + ... + 10m + 10 = 5m (m + 1) + 10

    LHS = RHS

    Hence, f (n) is valid for n = m + 1

    Thus,

    f (n) is valid for all positive integers
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