Question

Tính các tổng sau:

a)1+2+3+4+...+n

b)1+3+5+7+...+(2.n+1)

c)2+4+6+8+...+2.n

Answer 1

a) 1 + 2 + 3 + ... + n

= \(\frac{\left(n+1\right).n}{2}\)

b) 1 + 3 + 5 + 7 + ... + (2n + 1)

= \(\left(2n+1+1\right).\left(\frac{2n+1-1}{2}+1\right):2\)

\(=\left(2n+2\right).\left(\frac{2n}{2}+1\right):2\)

\(=2.\left(n+1\right).\left(n+1\right):2\)

\(=\left(n+1\right)^2\)

c) 2 + 4 + 6 + 8 + ... + 2.n

= 2.(1 + 2 + 3 + 4 + ... + n)

\(=2.\frac{\left(n+1\right).n}{2}\)

= (n + 1).n