Question

If $\underset{x\to 0^{+}}{lim}x(\left[\frac{1}{x}\right]+\left[\frac{5}{x}\right]+\left[\frac{11}{x}\right]+\left[\frac{19}{x}\right]+\left[\frac{29}{x}\right]+.......tonterms)=430$ (where [.] denotes the greatest integer function), then $n=$

• A. $8$
• B. $9$
• C. $10$
• D. $11$

Answer 1

The correct option is C $10$

${lim}_{x\to 0^{+}}x(\left[\frac{1}{x}\right]+\left[\frac{5}{x}\right]+\left[\frac{11}{x}\right]+\left[\frac{19}{x}\right]+\left[\frac{29}{x}\right]+.......tonterms)=430$

$⟹1+5+11+19+29+.........nterms=430$

$⟹1+5+11+19+29+.........+t_n=430$

$1+5+11+19+29+.........t_n$

By method of difference,

$t_n=n^2+n-1$

$\because S_n=\sum t_n$

$⟹\sum t_n=430$

$⟹\sum (n^2+n-1)=430$

$⟹\sum n^2+\sum n-\sum 1=430$

$⟹\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}-n=430$

$⟹\frac{n(n+1)}{2}[\frac{2n+1}{3}+1]-n=430$

$⟹\frac{n(n+1)}{2}\left[\frac{2n+4}{3}\right]-n=430$

$⟹\frac{n(n+1)(n+2)}{3}-n=430$

$⟹n(n+1)(n+2)-3n=430\times 3$

$⟹n^3+3n^2+2n-3n=1290$

$⟹n(n^2+3n-1)=10\times 129$

$⟹n(n^2+3n-1)=10\times ({10}^2+30-1)$

$\therefore n=10$

Option C is correct.