Find the $n$ th term of the series $3 + 7 x + 13 x^{2} + 21 x^{3} + . . . . .$

T_{n} = (n^{2} + n + 2) x^{n - 1}

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T_{n} = (n^{2} - n + 2) x^{n}

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T_{n} = (n^{2} + n + 1) x^{n - 1}

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T_{n} = (n^{2} + n + 2) x^{n}

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Solution

The correct option is D $T_{n} = (n^{2} + n + 1) x^{n - 1}$
The first factors of the terms of the series are as follows, $3, 7, 13, 21, . . .$
The difference of a term and its previous term, as follows:
$4, 6, 8, . . .$
$2, 2, . . .$
Here we see that the difference is same after two operations.
Hence the $n$ th term is of the form,
$u_{n} = a n^{2} + b n + c$
where $a, b, c$ are constants.
For $n = 1, 2, 3$
$3 = a + b + c \dots (i)$
$7 = 4 a + 2 b + c \dots (i i)$
$13 = 9 a + 3 b + c \dots (i i i)$
Solving $(i), (i i) a n d (i i i)$ , we get
$a = 1, b = 1, c = 1$
$∴ u_{n} = (n^{2} + n + 1)$
Hence $n$ th term of the given series is $T_{n} = (n^{2} + n + 1) x^{n - 1}$

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T_{n} = n^{2} + 4

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n

(1^{2} - t_{1}) + (2^{2} - t_{2}) + . . . . . . . + (n^{2} - t_{n}) = \frac{1}{3} n (n^{2} - 1)

t_{n}

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m a x [t_{n} - t_{k}]

ϵ [2, 10)

n > k

S_{n} = n \sum k = 1 \frac{n}{n^{2} + k n + k^{2}}

T_{n} = n - 1 \sum k = 0 \frac{n}{n^{2} + k n + k^{2}}

n = 1, 2, 3, . .

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