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(C) Let the series be $a_n = 5, 14, 29, 50, 77, \ldots$Calculate the first differences: $14-5=9, 29-14=15, 50-29=21, 77-50=27$.The differences are $9,

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$3n^2 + 2$

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(C) Let the series be $a_n = 5, 14, 29, 50, 77, \ldots$Calculate the first differences: $14-5=9, 29-14=15, 50-29=21, 77-50=27$.The differences are $9, 15, 21, 27, \ldots$,which form an arithmetic progression with a common difference of $6$.Since the second difference is constant $(6)$,the $n^{th}$ term is a quadratic expression of the form $a_n = An^2 + Bn + C$.For $n=1: A + B + C = 5$For $n=2: 4A + 2B + C = 14$For $n=3: 9A + 3B + C = 29$Subtracting equations: $(4A+2B+C) - (A+B+C) = 14-5 \implies 3A + B = 9$.$(9A+3B+C) - (4A+2B+C) = 29-14 \implies 5A + B = 15$.Subtracting these results: $(5A+B) - (3A+B) = 15-9 \implies 2A = 6 \implies A = 3$.Substituting $A=3$ into $3A+B=9$: $3(3) + B = 9 \implies B = 0$.Substituting $A=3, B=0$ into $A+B+C=5$: $3 + 0 + C = 5 \implies C = 2$.Thus,the $n^{th}$ term is $a_n = 3n^2 + 2$.

Find the $n^{th}$ term of the following series: $5, 14, 29, 50, 77, \ldots$

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