1/ \(=3^n.3^2+3^n=3^n\left(3^2+1\right)=10.3^n⋮10\)
2/ \(100.x+\left(1+2+3+...+100\right)=7450\)
Đến đây bạn tự làm nốt nhé
1. Ta có: \(3^{n+2}+3^n=3^n.\left(3^2+1\right)=3^n.\left(9+1\right)=3^n.10⋮10\)( đpcm )
2. \(\left(x+1\right)+\left(x+2\right)+.......+\left(x+100\right)=7450\)
\(\Leftrightarrow x+1+x+2+........+x+100=7450\)
\(\Leftrightarrow100x+\frac{100.101}{2}=7450\)
\(\Leftrightarrow100x+5050=7450\)
\(\Leftrightarrow100x=2400\)\(\Leftrightarrow x=24\)
Vậy \(x=24\)
1. 3n+2 + 3n = 3n( 32 + 1 ) = 3n.10 \(⋮\)10 \(\forall n\inℕ\)( đpcm )
2. ( x + 1 ) + ( x + 2 ) + ... + ( x + 100 ) = 7450
<=> ( x + x + ... + x ) + ( 1 + 2 + ... + 100 ) = 7450
<=> 100x + \(\frac{\left(100+1\right)\left[\left(100-1\right):1+1\right]}{2}\)= 7450
<=> 100x + 5050 = 7450
<=> 100x = 2400
<=> x = 24