Question

1;tinh

a;1-2-3+4+5-6-7+8+....+1993-1994

b;1^2-2^2+3^2-4^2+.....+99^2-100^2+101^2

Answer 1

1 - 2 - 3 + 4 + 5 - 6 - 7 + 8+ ... + 1993 - 1994

= ( 1 - 2 - 3 + 4 ) = ( 5 - 6 - 7 + 8 ) + ... + 1993 - 1994

= 0 + 0 + ... + 1993 - 1994 

= 0 + ( -1 ) = -1

b) ta có 1^2+2^2+...+n^2 = n(n+1)(2n+1)/6 
=>2^2+4^2+...+(2n)^2= 2^2(1^2+2^2+...+n^2)= 2n(n+1)(2n+1)/3 
và 1^2+2^2+...+(2n+1)^2=(2n+1)(2n+2)(4n+3)/... 
=>1^2+3^2+5^2+...+(2n+1)^2 = (2n+1)(2n+2)(4n+3)/6 - 2n(n+1)(2n+1)/3 = (2n+1)(n+1)(2n+3)/3
=>1^2-2^2+3^2-4^2+..... -(2n)^2+(2n+1)^2 = (2n+1)(n+1)(2n+3)/3 - 2n(n+1)(2n+1)/3 = (n+1)(2n+1) 
do đó ta có khi n = 100 thì 
1^2-2^2+3^2-4^2.....+99^2-100^2+101^2 = (100+1)*(2*100+1)=201*101

Mình cũng không chắc câu b cho lắm