Home -> Solved problems -> Challenging problem Challenging problem Solution Let’s start, we have an+1=2n−3an=2n−3×2n−1+32an−1=⋯=2n+∑j=1n(−3)j2n−j+(−3)n+1a0=2n+15+(a0−15)(−3)n+1 If a0=15, then we have no problem. If a0≠15, (23)n goes to 0 when n→∞, so an3n will have the same sign as (a0−15)(−1)n when n is large. Hence, an<an+1 will not hold then contradiction Home -> Solved problems -> Challenging problem Related Topics Find the volume of the square pyramid as a function of a and H by slicing method. Solution Prove that limx→0sinxx=1 Solution Prove that Solution Calculate the half derivative of x Solution Prove Wallis Product Using Integration Solution Calculate the radius R Solution Calculate the volume of Torus using cylindrical shells Solution Find the derivative of exponential x from first principles Solution Calculate the sum of areas of the three squares Solution Find the equation of the curve formed by a cable suspended between two points at the same height Solution Solve the equation for real values of x Solution Solve the equation for xϵR Solution Determine the angle x Solution Calculate the following limit Solution Calculate the following limit Solution Calculate the integral Solution Challenging problem Solution Home -> Solved problems -> Challenging problem Share the solution: Challenging problem