Question

Tính tổng sau

1) B= 1.2+2.3+3.4+......+99.100

2) C= \(1^2+2^2+3^2+...+99^2\)

3) D= \(\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right).....\left(1-\frac{1}{n^2}\right)\)

4) E=\(\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+....+\frac{1}{3^{100}}\)

Answer 1

B= 333300

C=328350

D=(n+1) /( n nhân 2)

E=(1/3 trừ 1/3^100):2

Answer 2

1)=>3B=1.2.3+2.3.3+3.4.3+...+99.100.3

3B=1.2.3+2.3.(4-1)+3.4.(5-2)+...+99.100.(101-98)

3B=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+99.100.101-98.99.100

3B=99.100.101

=>B=333300