Ck=ak+bk
where ak=a1+(k−1)(−3) and bk=b12k−1
Ck=a1+(k−1)(−3)+b12k−1
k=1∑10ck=k=1∑10ak+k=1∑10bk
C2=a1−3+b1(2)=12(1)....
C3=a1−6+4b1=13(2)......
by solving (1) & (2)
2b1−3=1⇒b1=2 and a1=11
k=1∑10ck=210(22+9(−3))+2−12(210−1)
=−25+2046=2021