Bài tập nâng cao · Bài 19
Bài tập nâng cao Cấp số nhân · Bài 19
Tính tổng $B=7+77+777+\cdots+\underbrace{777\ldots 7}_{n}$
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Lời giải
$B=7+77+777+\cdots+\underbrace{777\ldots 7}_{n}$
$\Leftrightarrow B=7\left(1+11+111+\cdots+\underbrace{111\ldots 1}_{n}\right)$
$\Leftrightarrow \frac{B}{7}=1+11+111+\cdots+\underbrace{111\ldots 1}_{n}$
$\Leftrightarrow \frac{9B}{7}=9+99+999+\cdots+\underbrace{99\ldots 9}_{n}$
$\Leftrightarrow \frac{9B}{7}=\left(10-1\right)+\left(10^{2}-1\right)+\left(10^{3}-1\right)+\cdots+\left(10^{n}-1\right)$
$\Leftrightarrow \frac{9B}{7}=\left(10+10^{2}+10^{3}+\cdots+10^{n}\right)-\underbrace{\left(1+1+\cdots+1\right)}_{n}$
$\Leftrightarrow \frac{9B}{7}=\frac{10\left(1-10^{n}\right)}{1-10}-n$
$\Leftrightarrow \frac{9B}{7}=\frac{10^{n+1}-9n-10}{9}$
$\Leftrightarrow B=\frac{7\left(10^{n+1}-9n-10\right)}{81}$
$\Leftrightarrow B=7\left(1+11+111+\cdots+\underbrace{111\ldots 1}_{n}\right)$
$\Leftrightarrow \frac{B}{7}=1+11+111+\cdots+\underbrace{111\ldots 1}_{n}$
$\Leftrightarrow \frac{9B}{7}=9+99+999+\cdots+\underbrace{99\ldots 9}_{n}$
$\Leftrightarrow \frac{9B}{7}=\left(10-1\right)+\left(10^{2}-1\right)+\left(10^{3}-1\right)+\cdots+\left(10^{n}-1\right)$
$\Leftrightarrow \frac{9B}{7}=\left(10+10^{2}+10^{3}+\cdots+10^{n}\right)-\underbrace{\left(1+1+\cdots+1\right)}_{n}$
$\Leftrightarrow \frac{9B}{7}=\frac{10\left(1-10^{n}\right)}{1-10}-n$
$\Leftrightarrow \frac{9B}{7}=\frac{10^{n+1}-9n-10}{9}$
$\Leftrightarrow B=\frac{7\left(10^{n+1}-9n-10\right)}{81}$