Question

Tính các tổng sau:

A=1.2+2.3+...+n.(n+1)+...+98.99

B=1.99+2.98+...+n(100-n)+...+98.2+99.1

Answer 1

a) 3.A = 1.2.3 + 2.3.(4 - 1) + 3.4.(5- 2) +...+n.(n+1).(n+2 - (n-1)) + ...+ 97.98.(99- 96) + 98.99.(100 - 97)

=> 3.A = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 +...+ 97.98.99 - 96.97.98 + 98.99.100 - 97.98.99

= 98.99.100

=> A = 98.99.100 : 3 = 323400

b) B gồm 99 số 1; 98 số 2;..; 2 số 98; 1 số 99

Có thể Viết lại B = 1 + (1+2) + (1+2+3) +...+ (1+2+3+...+98 + 99)

 =  \(\frac{1.2}{2}+\frac{2.3}{2}+\frac{3.4}{2}+...+\frac{98.99}{2}=\frac{1.2+2.3+3.4+...98.99}{2}=\frac{A}{2}=\frac{323400}{2}=161700\)

Answer 2

A = 1.2 + 2.3 + 3.4 +......+ 98.99

=> 3A = 1.2.3 + 2.3.(4 - 1) + 3.4.(5 - 2) + ........ + 98.99.(100 - 97)

=> 3A = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ........ + 98.99.100 - 97.98.99

=> 3A = (1.2.3 + 2.3.4 + 3.4.5 + ....... + 98.99.100) - (1.2.3 + 2.3.4 + ..... + 97.98.99)

=> 3A =  98.99.100

=> A = \(\frac{98.99.100}{3}=323400\)