This is from the chapter geometric progression.
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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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