Question

Prove that

∑_(i=5)^20▒〖3.2^i 〗= ∑_(k=0)^15▒〖96.2^k 〗

Answer 1

Use calculator

$$\displaystyle\sum_{i=5}^{20}(3.2)^i\approx18438554032.6177$$

$$\displaystyle\sum_{k=0}^{15}(96.2)^k\approx5.65150614585858⋅10^{29}$$

Therefore the statement is false.