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Question

Find the sum of the following series.
(i) 1+2+3+....+45
(ii) 162+172+182+....+252
(iii) 2+4+6+....+100
(iv) 7+14+21+....+490
(v) 52+72+92+392
(vi) 163+173+.....+353

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Solution

(i) 1+2+3..+45=45(45+1)2=1035
(ii) 162+172..+252=12+22+32...252(12+22...152)=25(26)(51)615(16)(31)6=55251240=4285
(iii) 2+4+6..+100=2(1+2+3...+50)=2×50(51)2=2550
(iv) 7+14+...+490=7(1+2+3...+70)=7×70(71)2=17395
(v) 52+72+92+392=1676
(vi) 163+173...+353=13+23+...353(13+23+33...+153)=(35(36)2)2(15(16)2)2=39690014400=382500

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