`int(sum_r=0^oo(x^r 3^r)/(r !))dx`

int (overset(oo)underset(r=o)sum(x^(r)2^(r))/(r!))dx=

int (sum_(r = 0)^(infty) (x^(r) 3^(r))/(r!) ) dx =

int(sum_(r=0)^(oo)(x^(r)backslash2^(r))/(r!))dx=(A)e^(x)+C(B)(e^(2x))/(2)+C(C)-(2)/(1-2x)+C(D)2e^(2x)+C

Find sum_(r=1)^oo (-1/3)^r

If f(x)=(4+x)^(n),"n" epsilonN and f^(r)(0) represents the r^(th) derivative of f(x) at x=0 , then the value of sum_(r=0)^(oo)((f^(r)(0)))/(r!) is equal to

If f(x)=(4+x)^(n),"n" epsilonN and f^(r)(0) represents the r^(th) derivative of f(x) at x=0 , then the value of sum_(r=0)^(oo)((f^(r)(0)))/(r!) is equal to

Let A={x in R:[x+3]+[x+4]<=3} and B={x in R:3^(x)(sum_(r=1)^(oo)(3)/(10^(r)))^(x-3)<3^(-3x)}

Find: sum_(r=1)^oo (-1/3)^r

sum_(r=0)^(oo)((r+1)((2)/(3))^(r)) is

int_(0)^(oo)(dx)/((x+r)^(3/2))