\(B=3+3^2-3^3+3^4-3^5+...-3^{99}+3^{100}\)
\(3B=3^2+3^3-3^4+3^5-3^6+.....-3^{100}+3^{101}\)
\(=>3B+B=\left(3^2+3^3-3^4+3^5-3^6+....-3^{100}+3 ^{101}\right)+\left(3+3^2-3^3+3^4-3^5+...-3^{99}+3^{100}\right)\)
\(=>4B=3+3^{101}+3^2+3^2=3+3^{101}+18=>B=\frac{3+18+3^{101}}{4}\)