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Question

Find the sum of the following series
(i) 26+27+28+.....+60
(ii) 1+3+5+... to 25 terms
(iii) 31+33+....+53

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Solution

(i) We have 26+27+28+....+60=(1+2+3+....+60)(1+2+3+....+25)
=601n251n
=60(60+1)225(25+1)2
=30×61)(25×13)=1830325=1505.
(ii) Here, n=25
1+3+5+.... to 25 terms=252(nk=1(2k1)=n2)
=625.
(iii) 31+33+....+53
=(1+3+5+....+53)(1+3+5+....+29)
=(53+12)2(29+12)2(1+3+5+....+l=(l+12)2)
=272152=504.

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