ii) 1+4-7+10+13-16+19+22 – 25+... to 3n terms.
Find the sum of 20 terms
of the series whose rth term is 3r +1.
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2+8+26+80+...+3^r-1 is the sum of the series, which can be written (S(3^r))-r where S denotes sum.
The GP is 3 3^2 3^3 etc. In this series the sum has a=3 and r=3 so the sum is a(r^n-1)/(r-1)=3(3^n-1)/2. But we need to subtract n from this: (3(3^n-1)/2)-n. Put n=4: 120-4=116=2+8+26+80, which is correct.
The second part is the sum of two GPs:
2(2^n-1)+3(3^n-1)/2. Put n=4: 30+120=150; (2+3)+(4+9)+(8+27)+(16+81)=5+13+35+97=150.
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