Given s and s, where s:1-1+1-1+1+...= t1 1 sa 1-2+2-2+... then. 1+2 (a) s, is true but s, is false (b) s is false but s, is true (c) both s and s, are true

option a is false because if S1=1-1+1-1+...=1=1 , since S1=1-(1-1+1-...)=1 Again S1=(1-1)+(1-1)+...=0 so option a is false . Option b is correct because we know that S2=1-2+2^2-2^...

A S1 xn=(-1)^n if n is even then S1is 1 if n is odd then S1 is -1 S2 (1+x)^-1 =1-x+x²-x³+...... if |x|<1 so S2 wrong