Answer to Question #20250 in Calculus for kawalpreet kaur
we can differentiate both parts
we use that e^t=sum( i=0 to infinity) t^n/n!
so we have
sum( i=0 to infinity) (ax)^n/n!+bx-3x-16sum(i=0 toinfinity) (4x)^n/2=0 sum( i=0 to infinity) (a^n-16*4^n)*x^n/n!=(3-b)x but it possible
if and only if for n=1 (a^n-16*4^n)=3-band for others (a^n-16*4^n)/n!=0 so a^n-16*4^n=0 for n>1 and n=0 but it mean
a=4*16^(1/n) for n>1 its impossible. there is no such a and b
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