\(B=1.4+2.5+3.6+...+99.102\)
\(=1.\left(2+2\right)+2.\left(2+3\right)+3.\left(2+4\right)+...+99.\left(2+100\right)\)
\(=1.2+2.1+2.3+2.2+3.4+2.3+...+99.100+2.99\)
\(=\left(1.2+2.3+...+99.100\right)+\left(2.1+2.2+2.3+...+2.99\right)\)
\(=333300+2.\left(1+2+3+...+99\right)\)
\(=333300+2.\left(\frac{99.100}{2}\right)\)
\(=333300+99.100=333300+9900=343200\)
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