Find the sum of the following series
(i) $26 + 27 + 28 + . . . . . + 60$
(ii) $1 + 3 + 5 + . . . t o 25 t e r m s$
(iii) $31 + 33 + . . . . + 53$

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Solution

(i) We have $26 + 27 + 28 + . . . . + 60 = (1 + 2 + 3 + . . . . + 60) - (1 + 2 + 3 + . . . . + 25)$
$= 60 \sum 1 n - 25 \sum 1 n$
$= \frac{60 (60 + 1)}{2} - \frac{25 (25 + 1)}{2}$
$= 30 \times 61) - (25 \times 13) = 1830 - 325 = 1505$ .
(ii) Here, $n = 25$
$∴ 1 + 3 + 5 + . . . . t o 25 t e r m s = 25^{2} (n \sum k = 1 (2 k - 1) = n^{2})$
$= 625$ .
(iii) $31 + 33 + . . . . + 53$
$= (1 + 3 + 5 + . . . . + 53) - (1 + 3 + 5 + . . . . + 29)$
$= {(\frac{53 + 1}{2})}^{2} - {(\frac{29 + 1}{2})}^{2} (1 + 3 + 5 + . . . . + l = {(\frac{l + 1}{2})}^{2})$
$= 27^{2} - 15^{2} = 504$ .

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