4 + 6 + 9 + 13 + 18 + ...

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Solution

Let $T_{n}$ be the nth term and $S_{n}$ be the sum of n terms of the given series.

Thus, we have:

$S_{n} = 4 + 6 + 9 + 13 + 18 + . . . + T_{n - 1} + T_{n}$ ...(1)

Equation (1) can be rewritten as:

$S_{n} = 4 + 6 + 9 + 13 + 18 + . . . + T_{n - 1} + T_{n}$ ...(2)

On subtracting (2) from (1), we get:

$S_{n} = 4 + 6 + 9 + 13 + 18 + . . . + T_{n - 1} + T_{n} S_{n} = 4 + 6 + 9 + 13 + 18 + . . . + T_{n - 1} + T_{n} 0 = 4 + [2 + 3 + 4 + 5 + 6 + . . . + (T_{n} - T_{n - 1})] - T_{n}$

The sequence of difference between successive terms is 2, 3, 4, 5,...

We observe that it is an AP with common difference 1 and first term 2.

Now,

$4 + [\frac{(n - 1)}{2} \{4 + (n - 2) 1\}] - T_{n} = 0 \Rightarrow 4 + [\frac{(n - 1)}{2} (n + 2)] - T_{n} = 0 \Rightarrow 4 + [\frac{n^{2} + n}{2} - 1] - T_{n} = 0 \Rightarrow [\frac{n^{2}}{2} + \frac{n}{2} + 3] = T_{n}$

$∵ S_{n} = \sum_{k = 1}^{n} T_{k} ∴ S_{n} = \sum_{k = 1}^{n} (\frac{k^{2}}{2} + \frac{k}{2} + 3) = \frac{1}{2} \sum_{k = 1}^{n} k^{2} + \frac{1}{2} \sum_{k = 1}^{n} k + \sum_{k = 1}^{n} 3 = \frac{n (n + 1) (2 n + 1)}{2 \times 6} + \frac{n (n + 1)}{2 \times 2} + 3 n = n (\frac{2 n^{2} + 3 n + 1 + 3 n + 3 + 36}{12}) = \frac{n}{12} (2 n^{2} + 6 n + 40) = \frac{n}{6} (n^{2} + 3 n + 20)$

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Similar questions

Verify the following:

(i) $18 \times [9 + (- 7)] = 18 \times 9 + 18 \times (- 7)$

(ii) $(- 13) \times [(- 6) + (- 19)] = (- 13) \times (- 6) + (- 13) \times (- 19)$

4 + 6 + 9 + 13 + 18 + . . . . . . n

\frac{9}{3} \div \frac{4}{6} \times \frac{13}{18} - \frac{3}{5} =

x \frac{13}{20}

18 : 9, 16 : 13

6 : 9

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