MCQ
$\sum_{r=1}^{20}\left(r^{2}+1\right)(r !)$ is equal to:
- A$22\,!-21 !$
- ✓$22\, !-2(21 \,!)$
- C$21\, !-2 (20\,!)$
- D$21 \,!-20\, !$
$\sum_{x=1}^{20}\left((r+1)^{2}-2 r\right) r !$
$\sum_{x=1}^{20}((r+1)(r+1) !-r \cdot r !)-\sum_{r=1}^{20} r \cdot r !$
$\sum_{x=1}^{20}((r+1)(r+1) !-r \cdot r !)-\sum_{r=1}^{20}((r+1) !-r !)$
$=(21.21-1)-(\lfloor 21-1)$
$=20.21 !=22 !-2 \cdot 21 !$
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($1$)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 ($2$)