Question

write out the sum.
\\(\sum\limits_{k=0}^{n-1} \frac{1}{4^{k+1}}\\)

find the first, second, third and last terms of the sum.
\\(\sum\limits_{k=0}^{n-1} \frac{1}{4^{k+1}} = \square + \square + \square + \dots + \square\\)

Explanation:

Step1: Find first term (k=0)

Substitute k=0: $\frac{1}{4^{0+1}} = \frac{1}{4}$

Step2: Find second term (k=1)

Substitute k=1: $\frac{1}{4^{1+1}} = \frac{1}{16}$

Step3: Find third term (k=2)

Substitute k=2: $\frac{1}{4^{2+1}} = \frac{1}{64}$

Step4: Find last term (k=n-1)

Substitute k=n-1: $\frac{1}{4^{(n-1)+1}} = \frac{1}{4^n}$

Answer:

$\frac{1}{4} + \frac{1}{16} + \frac{1}{64} + \dots + \frac{1}{4^n}$