How do you find the sum of #11+4i# from i=1 to 45?

1 Answer
Jan 27, 2017

Sum of #11+4i# from #i=1# to #45# is #4635#

Explanation:

#11+4i#, where #i=1# to #45#, basically represents

an arithmetic series, whose first term is #11+4xx1=15# and common difference #d=4#, as subsequent terms go up by #4#

Hence sum of the series for #n# terms is given by

#S_n=n/2(2a+(n-1)d)#

and the sum of #11+4i# from #i=1# to #45# is

#S_(45)=45/2(2xx15+(45-1)xx4)#

= #45/2xx(30+44xx4)#

= #45/2(30+176)#

= #45/2xx206=4635#