Answer to Question #217590 in Algebra for Bhatti

Question

Prove that 

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

Answer 1

$\displaystyle{\sum_{i=5}^{20} }3.2^i=\displaystyle{\sum_{i=0}^{20} }3.2^i-\displaystyle{\sum_{i=0}^{4} }3.2^i$

$\displaystyle{\sum_{i=0}^{20} }3.2^i=\frac{1-3.2^{21}}{1-3.2}$

$\displaystyle{\sum_{i=0}^{4} }3.2^i=\frac{1-3.2^{5}}{1-3.2}$

$\displaystyle{\sum_{k=0}^{15} }96.2^k=\frac{1-96.2^{16}}{1-96.2}$

$\displaystyle{\sum_{i=5}^{20} }3.2^i=18438554032.61773674904295571456$

$\displaystyle{\sum_{k=0}^{15} }96.2^k=565150614585858182318186506187.03$