MCQ
$\sum_{r=1}^{20}\left(r^{2}+1\right)(r !)$ ની કિમંત મેળવો.
- 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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વિધાન $-2:$ રેખા $y = mx - \frac{1}{{2m}}(m \ne 0)$ પરવલય $y^2 = - 2x$ ના બિંદુ $\left( { - \frac{1}{{2{m^2}}}, - \frac{1}{m}} \right)$ આગળના સ્પર્શકનું સમીકરણ છે