assume true for n=p

then,

$\displaystyle S_p=(0.9)^p(1200)+1000(1-0.9^p)$

$\displaystyle S_{p+1}=(0.9)^p(1200)+1000(1-0.9^p)+100-0.1[(0.9)^p(1200)+1000(1-0.9^p)] $

$\displaystyle S_{p+1}=(0.9)^p(1200)+1000(1-0.9^p) +100 -(1-0.9)(0.9^p(1200))-1000(1-0.9)(1-0.9^p)$

$\displaystyle S_{p+1}=(0.9)^p(1200)+1000(1-0.9^p) +100 -0.9^p(1200)+0.9^{p+1}(1200)-1000((1-0.9^p)-0.9+0.9^{p+1})$

$\displaystyle S_{p+1}=1000(1-0.9^p)+100 +0.9^{p+1}(1200)-1000(1-0.9^p)+900-1000(0.9^{p+1})$

$\displaystyle S_{p+1}=100 +0.9^{p+1}(1200)+900-1000(0.9^{p+1})$

$\displaystyle S_{p+1}=0.9^{p+1}(1200)+1000-1000(0.9^{p+1})$

$\displaystyle S_{p+1}=0.9^{p+1}(1200)+1000(1-0.9^{p+1})$

so the statement is true for n=p+1