Find the n^{th} term of the following series: | vedclass

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(B) Let the series be $a_n = 1, 7, 17, 31, 49, \ldots$The differences between consecutive terms are:$7 - 1 = 6$$17 - 7 = 10$$31 - 17 = 14$$49 - 31 = 1

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$2n^2 - 1$

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(B) Let the series be $a_n = 1, 7, 17, 31, 49, \ldots$The differences between consecutive terms are:$7 - 1 = 6$$17 - 7 = 10$$31 - 17 = 14$$49 - 31 = 18$The sequence of differences is $6, 10, 14, 18, \ldots$,which is an Arithmetic Progression with first term $a = 6$ and common difference $d = 4$.The $n^{th}$ term of the original series can be expressed as a quadratic form $a_n = An^2 + Bn + C$.For $n=1: A + B + C = 1$For $n=2: 4A + 2B + C = 7$For $n=3: 9A + 3B + C = 17$Subtracting the first from the second: $3A + B = 6$Subtracting the second from the third: $5A + B = 10$Subtracting these results: $2A = 4 \implies A = 2$.Substituting $A=2$ into $3A + B = 6$,we get $6 + B = 6 \implies B = 0$.Substituting $A=2, B=0$ into $A + B + C = 1$,we get $2 + 0 + C = 1 \implies C = -1$.Thus,the $n^{th}$ term is $a_n = 2n^2 - 1$.

Find the $n^{th}$ term of the following series:
$1, 7, 17, 31, 49, \ldots$

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Find the $n^{th}$ term of the following series:$1, 7, 17, 31, 49, \ldots$