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)6−15(16)(31)6=5525−1240=4285