Question

Tìm n ∈ ℕ^*

^{1) 1 + 2 + 3 + ..... + n = 4950}

^{2) 2 + 4 + 6 + 8 + ..... + ( 2n + 2 ) = 4950}

Answer 1

1) \(1+2+3...............+n=4950\)

\(\Leftrightarrow\dfrac{n\left(n+1\right)}{2}=4950\)

\(\Leftrightarrow n\left(n+1\right)=4950.2\)

\(\Leftrightarrow n\left(n+1\right)=9900\)

\(\Leftrightarrow n\left(n+1\right)=99.100\)

\(\Leftrightarrow n=99\left(tm\right)\)

Vậy ................

Answer 2

1)

1 + 2 + 3 + ... + n = 4950

(n + 1) . n : 2 = 4950

(n + 1) . n = 9900

(n + 1) . n = 100 . 99

Vậy n = 99