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Giải:
a, \(B=1^2+2^2+3^2+...+99^2+100^2.\)
\(B=1\left(2-1\right)+2\left(3-1\right)+3\left(4-1\right)+...+99\left(100-1\right)+100\left(101-1\right).\)
\(B=1.2-1.1+2.3-1.2+3.4-1.3+...+99.100-1.99+100.101-1.100.\)
\(B=\left(1.2+2.3+3.4+...+99.100+100.101\right)-\left(1+2+3+...+100\right).\)
\(B=\dfrac{\left[1.2.3+2.3\left(4-1\right)+3.4\left(5-2\right)+...+100.101\left(102-99\right)\right]}{3}+\dfrac{100\left(100+1\right)}{2}.\)
\(B=\dfrac{\left(1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+100.101.102-99.100.101\right)}{3}+5050.\)
\(B=\dfrac{100.101.102}{3}+5050.\)
\(B=343400+5050=348450.\)
Vậy \(B=348450.\)
\(C=...\) (làm tương tự con \(B\)).
\(D=...\) (hình như đề sai).
\(T=1.100+2.99+3.98+...+99.2+100.1.\)
\(T=1.100+2.\left(100-1\right)+3.\left(100-2\right)+...+99\left(100-98\right)+100\left(100-99\right).\)
\(T=1.100+100.2+1.2+100.3+2.3+...+100.99+98.99+100.100+99.100.\)
\(T=100\left(1+2+3+...+100\right)-\left(1.2+2.3+3.4+...+99.100\right).\)
\(T=100.\dfrac{100.101}{2}-\dfrac{99.100.101}{3}.\)
\(T=100.5050-333300.\)
\(T=505000-333300=171700.\)
Vậy \(T=171700.\)
\(S=1.2.3+2.3.4+3.4.5+...+98.99.100.\)
\(4S=4\left(1.2.3+2.3.4+3.4.5+...+98.99.100\right).\)
\(4S=1.2.3.4+2.3.4.4+3.4.5.4+...+98.99.100.4.\)
\(4S=1.2.3\left(5-1\right)+2.3.4\left(6-2\right)+...+98.99.100\left(101-97\right).\)
\(4S=1.2.3.4+2.3.4.5-1.2.3.4+3.4.5.6-2.3.4.5+...+98.99.100.101-97.98.99.100.\)
\(4S=\left(1.2.3.4-1.2.3.4\right)+\left(2.3.4.5-2.3.4.5\right)+...+\left(97.98.99.100-97.98.99.100\right)+98.99.100.101.\)
\(4S=0+0+...+0+98.99.100.101.\)
\(4S=98.99.100.101.\)
\(4S=97990200.\)
\(\Rightarrow S=\dfrac{97990200}{4}=24497550.\)
Vậy \(S=24497550.\)
~ Học tốt!!! ~
=1.100 + 2.(100-1) + 3.(100 - 2 ) +..... + 99 .(100 - 98) + 100 .(100 - 99 )
=1.100 + 2. 100 - 1.2 + 3.100 - 3.2 +...+ 99 . 100 - 99 . 98 + 100 .100 - 100.99
= 100 (1 + 2 + 3 +... +100) - (1.2 +3 .2 +...+ 99.98 + 100.99
=100.100.101 /2 - 99.100.101 / 3
= 505000 - 333300= 171700
Đ/S : 171700
=>F=100(1+2+3+...+100)-(1.2+2.3+3.4+...+99.100)
=>F=171700