AMC 8 Daily Practice - New Patterns

Complete problem set with solutions and individual problem pages

Problem 8 Medium

For natural numbers a and n, the operation a \Delta n is defined as a^{n} + a^{n-1}.For example, 3 \Delta 2 = 3^{2} + 3 = 12. Compute the sum:   1 \Delta 2 + 2 \Delta 2 + 3 \Delta 2 + \cdots + 99 \Delta 2.

  • A.

    4950

  • B.

    9900

  • C.

    328530

  • D.

    333300

  • E.

    100000

Answer:D

The expression can be expanded as:   1 \Delta 2 + 2 \Delta 2 + 3 \Delta 2 + \cdots + 99 \Delta 2 = 1^{2} + 1 + 2^{2} + 2 + 3^{2} + 3 + \cdots + 99^{2} + 99 = 1^{2} + 2^{2} + 3^{2} + \cdots + 99^{2} + 1 + 2 + 3 + \cdots + 99

Using the formulas:   1^{2} + 2^{2} + 3^{2} + \cdots + n^{2} = \frac{n(n+1)(2n+1)}{6}， 1 + 2 + 3 + \cdots + n = \frac{n(n+1)}{2}

Substitute n = 99 into the formulas:   1^{2} + 2^{2} + 3^{2} + \cdots + 99^{2} + 1 + 2 + 3 + \cdots + 99 = \frac{99 \cdot 100 \cdot 199}{6} + \frac{100 \cdot 99}{2}=328350 + 4950 = 333300

The final result: \boxed{333300}

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