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a:
Số số hạng trong dãy M là:
(1002-12):10+1=100(số)
=>Sẽ có 50 cặp (1002;992); (982;972);....;(22;12) có hiệu bằng 10
\(M=1002-992+982-972+...+22-12\)
\(=\left(1002-992\right)+\left(982-972\right)+...+\left(22-12\right)\)
\(=10+10+...+10\)
=10*50=500
b: \(N=\left(202+182+...+42+22\right)-\left(192+172+...+32+12\right)\)
\(=\left(202-192\right)+\left(182-172\right)+...+\left(22-12\right)\)
=10+10+...+10
=10*10=100
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2² + 4² + 6² + ... + 16² + 18²
= 4.(1 + 2² + 3² + ... + 8² + 9²)
= 4.285
= 1140
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Ta có \(2^2+4^2+...+20^2=2^2\left(1^2+2^2+...+10^2\right)=2^2.385=1540\).
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S = 22 + 42 + 62 + ... + 202
= (2.1)2 + (2.2)2 + (2.3)2 ... (2.10)2
= 22.12 + 22.22 + 22.32 + ... + 22.102
= 22 (12 + 22 + ... + 102 )
= 4 . 385 = 1540
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Ta có : \(1^2+2^2+3^2+.....+10^2=385\)
\(\Leftrightarrow2^2\left(1^2+2^2+3^2+.....+10^2\right)=2^2.385\)
\(\Leftrightarrow2^2+4^2+6^2+.....+20^2=4.385\)
\(\Leftrightarrow2^2+4^2+6^2+.....+20^2=1540\)
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\(B=1\cdot2\cdot3+2\cdot3\cdot4+...+\left(n-1\right)\cdot n\cdot\left(n+1\right)\)
=>\(4B=1\cdot2\cdot3\cdot4+2\cdot3\cdot4\cdot4+...+\left(n-1\right)\cdot n\left(n+1\right)\cdot4\)
=>\(4B=1\cdot2\cdot3\cdot4+2\cdot3\cdot4\left(5-1\right)+...+\left(n-1\right)\cdot n\left(n+1\right)\left[\left(n+2\right)-\left(n-2\right)\right]\)
=>\(4B=1\cdot2\cdot3\cdot4-1\cdot2\cdot3\cdot4+...+\left(n-2\right)\left(n-1\right)\cdot n\cdot\left(n+1\right)-\left(n-2\right)\cdot\left(n-1\right)\cdot n\cdot\left(n+1\right)+\left(n-1\right)\cdot n\left(n+1\right)\left(n+2\right)\)
=>\(4B=\left(n-1\right)\cdot n\cdot\left(n+1\right)\left(n+2\right)\)
=>\(B=\dfrac{\left(n-1\right)\cdot n\left(n+1\right)\left(n+2\right)}{4}\)
\(C=1\cdot4+2\cdot5+3\cdot6+...+n\left(n+3\right)\)
\(=1\cdot\left(1+3\right)+2\left(2+3\right)+...+n\left(n+3\right)\)
\(=\left(1^2+2^2+...+n^2\right)+3\left(1+2+...+n\right)\)
\(=\dfrac{n\left(n+1\right)\left(2n+1\right)}{6}+3\cdot\dfrac{n\left(n+1\right)}{2}\)
\(=\dfrac{n\left(n+1\right)\left(2n+1\right)}{6}+\dfrac{3n\left(n+1\right)}{2}\)
\(=\dfrac{n\left(n+1\right)}{2}\cdot\left(\dfrac{2n+1}{3}+3\right)\)
\(=\dfrac{n\left(n+1\right)}{2}\cdot\dfrac{2n+1+9}{3}\)
\(=\dfrac{n\left(n+1\right)\left(n+5\right)}{3}\)
\(D=1^2+2^2+...+n^2\)
\(=1+\left(1+1\right)\cdot2+\left(1+2\right)\cdot3+...+\left(1+n-1\right)\cdot n\)
\(=1+2+3+...+n+\left(1\cdot2+2\cdot3+...+\left(n-1\right)\cdot n\right)\)
Đặt \(A=1+2+3+...+n;E=1\cdot2+2\cdot3+...+\left(n-1\right)\cdot n\)
\(E=1\cdot2+2\cdot3+...+\left(n-1\right)\cdot n\)
=>\(3E=1\cdot2\cdot3+2\cdot3\cdot3+...+\left(n-1\right)\cdot n\cdot3\)
=>\(3E=1\cdot2\cdot3+2\cdot3\cdot\left(4-1\right)+...+\left(n-1\right)\cdot n\left[\left(n+1\right)-\left(n-2\right)\right]\)
=>\(3E=1\cdot2\cdot3-1\cdot2\cdot3+2\cdot3\cdot4+...+\left(n-1\right)\cdot n\left(n-2\right)-\left(n-1\right)\cdot n\left(n-2\right)+\left(n-1\right)\cdot n\cdot\left(n+1\right)\)
=>\(3E=\left(n-1\right)\cdot n\left(n+1\right)=n^3-n\)
=>\(E=\dfrac{n^3-n}{3}\)
\(A=1+2+3+...+n\)
Số số hạng là n-1+1=n(số)
Tổng của dãy số là: \(A=\dfrac{n\left(n+1\right)}{2}\)
=>\(D=\dfrac{n^3-n}{3}+\dfrac{n\left(n+1\right)}{2}\)
\(=\dfrac{2n^3-2n+3n^2+3n}{6}\)
=>\(D=\dfrac{2n^3+3n^2+n}{6}\)
\(\Rightarrow N=20^2-19^2+18^2-17^2+...+2^2-1^2\)
\(=\left(20+19\right)\left(20-19\right)+\left(18+17\right)\left(18-17\right)+...+\left(2+1\right)\left(2-1\right)\)
\(=20+19+18+17+...+2+1\)\(=\left(20+1\right)+\left(19+2\right)+...+\left(11+10\right)\)(có 10 cặp)
=21x10=210