Browse · MATH Print → jmc algebra senior Problem What is the value of the sum 32+3222+3323+…+310210? Express your answer as a common fraction. Solution — click to reveal This is the sum of the series a1+a2+…+a10 with a1=32 and r=32.Thus, S=1−ra(1−rn)=32⋅1−321−(32)10=32⋅311−590491024=32⋅13⋅5904958025=590492⋅58025=59049116050. Final answer \frac{116050}{59049} ← Previous problem Next problem →