Simplify #S_(k+1)# completely. Thanks?!!

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Answer 1 · 2018-02-06T15:05:24

#S_(k+1)=3/2k^2+5/2k+1#

We have an arithmetic series #1+4+7+...#, where common difference is #3#. It has its #n^(th)# term as #1+(n-1)3=3n-2#.

As #S_n# is sum of the series upto #n# terms and #S_n=n/2(3n-1)#

#S_k=k/2(3k-1)=3/2k^2-k/2#, sum of first #k# terms

and #S_(k+1)=(k+1)/2(3(k+1)-1)#

= #1/2(k+1)(3k+2)#

= #1/2(3k^2+2k+3k+2)#

= #3/2k^2+5/2k+1#, sum of first #k+1# terms