Find the $n^{th}$ term of the sequence 2, 10, 30, 68, 130 ...

5n + 20

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n^{2}

+ 1

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n^{3}

+ n

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n^{4}

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Solution

The correct option is C $n^{3}$ + n
$T_{2}$ = $2^{3}$ + 2 = 8+2 =10
$T_{3}$ = $3^{3}$ + 3 = 27+3 = 30
$T_{4}$ = $4^{3}$ + 4 = 64+4 = 68
$T_{5}$ = $5^{3}$ + 5 = 125+5 = 130

similarly $T_{n}$ = $n^{3}$ + n

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