The arithmetic sequence 2, 12, 36, 80.... is a function of

n^{2} (n - 1)

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n (n + 1)

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n^{2} (n + 1)

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n^{2} (n + 2)

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Solution

The correct option is D $n^{2} (n + 1)$
The arithmetic sequence 2, 12, 36, 80.... is a function of $n^{2} (n + 1)$ .
When n = 1, $1^{2} (1 + 1) = 2$
When n = 2, $2^{2} (2 + 1) = 12$
When n = 3, $3^{2} (3 + 1) = 36$
When n = 4, $4^{2} (4 + 1) = 80$ ....
So, $2, 12, 36, 80$ .. is the sequence for the function $n^{2} (n + 1)$ .

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Similar questions

Find the sum of first $n$ natural numbers.

1 \times n + 2 (n - 1) + 3 (n - 2) + \dots \dots + (n - 1) \times 2 + n \times 1

\frac{{1.2}^{2} + {2.3}^{2} + {3.4}^{2} + . \cdot . \cdot . \cdot . \cdot + n (n + 1)^{2}}{1^{2} .2 + 2^{2} .3 + 3^{2} .4 + + n^{2} (n + 1)} =

\frac{1 \times 2^{2} + 2 \times 3^{2} + 3 \times 4^{2} + \dots + n (n + 1)^{2}}{1^{2} \times 2 + 2^{2} \times 3 + 3^{2} \times 4 + \dots + n^{2} (n + 1)} = \frac{8}{7}

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\frac{{1.2}^{2} + {2.3}^{2} + {3.4}^{2} + . . . n (n + 1)^{2}}{1.2 + 2^{2} .3 + 3^{2} .4 + . . . n^{2} (n + 1)}

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