Here in given geometric series #a_1=4# and #r=-3#
Now if the first term of geometric series is #a# and common ratio is #r#, the sum of geometric series #S-n# is
#S_n=a+ar+ar^2+.................+ar^(n-1)# and therefore
#rS_n=ar+ar^2+.................+ar^(n-1)+ar^n#
and subtracting first from second #S_n(r-1)=a(r^n-1)#
and #S_n=(a(r^n-1))/(r-1)# if #r>1# or #S_n=(a(1-r^n))/(1-r)# if #r<1#
In the given geometric series #Sigma144(-1/2)^(n-1)#, the first term is #144# and common ratio is #-1/2#
Hence #S_7=(144*(1-(-1/2)^7))/(1-(-1/2))#
= #(144*(1-(-1/128)))/(1+1/2)#
=#144xx(129/128)/(3/2)#
= #48cancel144xx129/128xx2/(1cancel3)#
= #48xx129/128xx2#
= #3cancel48xx129/(4cancel8cancel128)xxcancel2#
= #387/4=96 3/4#
