Compute2+64100+2+2⋅6499+2+3⋅6498+⋯+2+98⋅643+2+99⋅642+2+100⋅64.
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