1
JEE Main 2021 (Online) 26th February Evening Shift
+4
-1
The sum of the series

$$\sum\limits_{n = 1}^\infty {{{{n^2} + 6n + 10} \over {(2n + 1)!}}}$$ is equal to :
A
$${{41} \over 8}e + {{19} \over 8}{e^{ - 1}} - 10$$
B
$${{41} \over 8}e - {{19} \over 8}{e^{ - 1}} - 10$$
C
$${{41} \over 8}e + {{19} \over 8}{e^{ - 1}} + 10$$
D
$$- {{41} \over 8}e + {{19} \over 8}{e^{ - 1}} - 10$$
2
JEE Main 2021 (Online) 26th February Morning Shift
+4
-1
The sum of the infinite series
$$1 + {2 \over 3} + {7 \over {{3^2}}} + {{12} \over {{3^3}}} + {{17} \over {{3^4}}} + {{22} \over {{3^5}}} + ......$$ is equal to :
A
$${9 \over 4}$$
B
$${13 \over 4}$$
C
$${15 \over 4}$$
D
$${11 \over 4}$$
3
JEE Main 2021 (Online) 26th February Morning Shift
+4
-1
In an increasing geometric series, the sum of the second and the sixth term is $${{25} \over 2}$$ and the product of the third and fifth term is 25. Then, the sum of 4th, 6th and 8th terms is equal to :
A
30
B
32
C
26
D
35
4
JEE Main 2021 (Online) 25th February Morning Slot
+4
-1
If $$0 < \theta ,\phi < {\pi \over 2},x = \sum\limits_{n = 0}^\infty {{{\cos }^{2n}}\theta } ,y = \sum\limits_{n = 0}^\infty {{{\sin }^{2n}}\phi }$$ and $$z = \sum\limits_{n = 0}^\infty {{{\cos }^{2n}}\theta .{{\sin }^{2n}}\phi }$$ then :
A
xy $$-$$ z = (x + y)z
B
xyz = 4
C
xy + z = (x + y)z
D
xy + yz + zx = z
JEE Main Subjects
Physics
Mechanics
Electricity
Optics
Modern Physics
Chemistry
Physical Chemistry
Inorganic Chemistry
Organic Chemistry
Mathematics
Algebra
Trigonometry
Coordinate Geometry
Calculus
EXAM MAP
Joint Entrance Examination