The nth term of the series 1+3+7+13+21+ldots | vedclass

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(A) Let the $n$th term of the series be $t_n$. The series is $1, 3, 7, 13, 21, \ldots$. Taking the differences between consecutive terms: $3-1=2, 7-3=

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$100$

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(A) Let the $n$th term of the series be $t_n$. The series is $1, 3, 7, 13, 21, \ldots$. Taking the differences between consecutive terms: $3-1=2, 7-3=4, 13-7=6, 21-13=8, \ldots$. This is an arithmetic progression of differences: $2, 4, 6, 8, \ldots$. The $n$th term $t_n$ can be expressed as $t_n = t_1 + \sum_{k=1}^{n-1} (2k)$. $t_n = 1 + 2 	imes \frac{(n-1)n}{2} = 1 + n(n-1) = n^2 - n + 1$. Given $t_n = 9901$,we have $n^2 - n + 1 = 9901$. $n^2 - n - 9900 = 0$. Factoring the quadratic equation: $n^2 - 100n + 99n - 9900 = 0$. $n(n-100) + 99(n-100) = 0$. $(n-100)(n+99) = 0$. Since $n$ must be a positive integer,$n = 100$.

The $n$th term of the series $1+3+7+13+21+\ldots$ is $9901$. The value of $n$ is

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