Question

2 Tìm n thuộc N thỏa mãn:

a, 1+2+3+...+n=231

b, 11+12+...+n=176

c, 1+3+5+...+(2n-1)=169

Answer 1

a) 1 + 2 + 3 + ... + n = 231

=> \(\frac{\left(1+n\right).n}{2}=231\)

=> (1 + n).n = 231.2

=> (1 + n).n = 462 = 21.22

=> n = 21

Vậy n = 21

b) 11 + 12 + ... + n = 176

=> \(\frac{11+n}{2}.\left(\frac{n-11}{1}+1\right)=176\)

=> (11 + n).(n - 10) = 176.2

=> (11 + n).(n - 10) = 352 = 32.11

=> n - 10 = 11; 11 + n = 32

=> n = 21

Vậy n = 21

c) 1 + 3 + 5 + ... + (2n - 1) = 169

\(\frac{\left(2n-1+1\right)}{2}.\left(\frac{2n-1-1}{2}+1\right)=169\)

=> \(\frac{2n}{2}.\left(\frac{2n-2}{2}+1\right)=169\)

=> n.(n - 1 + 1) = 169

=> n² = 169 = 13²

Vậy n = 13

Answer 2

a) n=21

b) n=21

c) n=13