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Ta có:
\(A=\left(1000-1^3\right).\left(1000-2^3\right).........\left(1000-50^3\right)\)
\(A=\left(1000-1^3\right).\left(1000-2^3\right).......\left(1000-10^3\right)........\left(1000-50^3\right)\)
\(A=0.\left(1000-1^3\right).\left(1000-2^3\right)....\left(1000-9^3\right).\left(1000-11^3\right).....\left(1000-50^3\right)\)
A = 0
(1000 - 13) x (1000 - 23) x (1000 - 33) x ... x (1000 - 103) x ... x (1000 - 503)
= (1000 - 13) x (1000 - 23) x (1000 - 33) x ... x (1000 - 1000) x ... x (1000 - 503)
= (1000 - 13) x (1000 - 23) x (1000 - 33) x ... x 0 x ... x (1000 - 503)
= 0
(1000 - 13)(1000 - 23)(1000 - 33)........(1000 - 503)
= (1000 - 13)(1000 - 23)(1000 - 33)....(1000 - 103)....(1000 - 503)
= (1000 - 13)(1000 - 23)(1000 - 33).....(1000 - 1000)...(1000 - 503)
= (1000 - 13)(1000 - 23)(1000 - 33).....0...(1000 - 503)
= 0
\(f\left(x\right)=x\left(x+1\right)+\left(x+1\right)\left(x+2\right)+\left(x+2\right)\left(x+3\right)+...+\left(x+49\right)\left(x+50\right)\)
\(f\left(1\right)=1\left(1+1\right)+\left(1+1\right)\left(1+2\right)+\left(1+2\right)\left(1+3\right)+...+\left(1+49\right)\left(1+50\right)\)
\(f\left(1\right)=1.2+2.3+3.4+...+50.51\)
Gọi f(1) = A
\(3A=1.2.3+2.3.3+3.4.3+...+50.51.3\)
\(3A=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+50.51.\left(52-49\right)\)
\(3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+50.51.52-49.50.51\)
\(3A=50.51.52\)=44200
Vậy f(1) = 44200
( 1000 - 13 ) . ( 1000 - 23 ) ... ( 1000 - 503 )
= ( 1000 - 13 ) . ( 1000 - 23 ) ... ( 1000 - 103 ) ... ( 1000 - 503 )
= ( 1000 - 13 ) . ( 1000 - 23 ) ... 0 ... ( 1000 - 503 )
= 0
a: \(=\dfrac{1}{50}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)\)
\(=\dfrac{1}{5}-\dfrac{49}{50}=\dfrac{-48}{50}=-\dfrac{24}{25}\)
501−(1−21+21−31+...+491−501)
=15−4950=−4850=−2425=51−5049=50−48=−2524