MCQ
The sum $\frac{3}{{{1^2}}} + \frac{5}{{{1^2} + {2^2}}} + \frac{7}{{{1^2} + {2^2} + {3^2}}} + ....$ upto $11-$ terms is
- A$\frac {7}{2}$
- B$\frac {11}{4}$
- ✓$\frac {11}{2}$
- D$\frac {60}{11}$
$\frac{3}{{12}} + \frac{5}{{{1^2} + {2^2}}} + \frac{7}{{{1^2} + {2^2} + {3^2}}} + .....$
${n^{th}}$ term $ = {T_n}$
$ = \frac{{2n + 1}}{{\frac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}}} = \frac{6}{{n\left( {n + 1} \right)}}$
or ${T_n} = 6\left[ {\frac{1}{n} - \frac{1}{{n + 1}}} \right]$
$\therefore {S_n} = $
$\sum {{T_n} = 6\sum {\frac{1}{n} - 6} \sum {\frac{1}{{n + 1}}} } = \frac{{6n}}{n} - \frac{6}{{n + 1}}$
$ = 6 - \frac{6}{{n + 1}} = \frac{{6n}}{{n + 1}}$
So, sum upto $11$ terms means
${S_{11}} = \frac{{6 \times 11}}{{11 + 1}} = \frac{{66}}{{12}} = \frac{{33}}{6} = \frac{{11}}{2}$
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$(A)$ $f(x)$ is monotonically increasing on $[1, \infty)$
$(B)$ $f(x)$ is monotonically decreasing on $(0,1)$
$(C)$ $f(x)+f\left(\frac{1}{x}\right)=0$, for all $x \in(0, \infty)$
$(D)$ $f\left(2^x\right)$ is an odd function of $x$ on $R$
($1$)The orthocentre of the triangle $F_1 M N$ is
($A$) $\left(-\frac{9}{10}, 0\right)$ ($B$) $\left(\frac{2}{3}, 0\right)$ ($C$) $\left(\frac{9}{10}, 0\right)$ ($D$) $\left(\frac{2}{3}, \sqrt{6}\right)$
($2$) If the tangents to the ellipse at $M$ and $N$ meet at $R$ and the normal to the parabola at $M$ meets the $x$-axis at $Q$, then the ratio of area of the triangle $M Q R$ to area of the quadrilateral $M F_{\mathrm{I}} N F_2$ is
($A$) $3: 4$ ($B$) $4: 5$ ($C$) $5: 8$ ($D$) $2: 3$
Givan the answer qestion ($1$) and ($2$)