Compute\(\frac{2 + 6}{4^{100}} + \frac{2 + 2 \cdot 6}{4^{99}} + \frac{2 + 3 \cdot 6}{4^{98}} + \dots + \frac{2 + 98 \cdot 6}{4^3} + \frac{2 + 99 \cdot 6}{4^2} + \frac{2 + 100 \cdot 6}{4}.\)
sumfor(n, 1, 100, (2 +6*n) /(4^(101 - n)))==It converges to 200
R==Common Ratio
R==99 / 400
SUM== 150.50 * [1 - (99/400)^100] / [1 - (99/400)], solve for SUM
SUM==200
+1162 
