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CBSE Questions for Class 11 Engineering Maths Sequences And Series Quiz 12 - MCQExams.com
CBSE
Class 11 Engineering Maths
Sequences And Series
Quiz 12
Report Question
0%
6
0%
7
0%
8
0%
9
0%
10
Find sum of series:
1.3.5
+
3.5.7
+
5.7.9.....
?
Report Question
0%
S
n
=
(
2
n
−
1
)
(
2
n
+
1
)
(
2
n
+
3
)
(
2
n
+
5
)
8
+
15
8
0%
S
n
=
(
2
n
−
1
)
(
2
n
+
1
)
(
2
n
+
3
)
(
2
n
+
5
)
18
+
15
18
0%
S
n
=
(
2
n
−
1
)
(
2
n
+
1
)
(
2
n
+
3
)
(
2
n
+
5
)
20
+
15
8
0%
S
n
=
(
2
n
−
1
)
(
2
n
+
1
)
(
2
n
+
3
)
(
2
n
+
5
)
6
+
1
8
sin
(
1
√
2
)
+
sin
(
√
2
−
1
√
6
)
+
⋯
+
sin
(
√
n
−
√
n
−
1
√
n
(
n
+
1
)
)
+
…
∞
=
Report Question
0%
π
0%
π
2
0%
π
4
0%
3
π
2
The sum of the series
1
+
1
1
!
.
1
4
+
1
⋅
3
2
!
(
1
4
)
2
+
1
⋅
3
⋅
5
3
!
(
1
4
)
3
+
.
.
.
.
.
.
.
.
to
∞
is ?
Report Question
0%
√
2
0%
2
0%
1
√
2
0%
√
3
The value of
(
21
C
1
−
10
C
1
)
+
(
21
C
2
−
10
C
2
)
+
(
3
C
1
−
10
C
3
)
+
(
21
C
4
−
10
C
4
)
+
.
.
.
.
.
(
21
C
10
−
10
C
10
)
is
Report Question
0%
2
20
−
2
10
0%
2
21
−
2
11
0%
2
21
−
2
10
0%
2
20
−
2
9
If
a
n
=
n
∑
r
=
0
1
n
C
r
, the value of
n
∑
r
=
0
n
−
2
r
n
C
r
Report Question
0%
n
2
a
n
0%
1
2
a
n
0%
n
a
n
0%
0
The value of
100
∑
r
=
2
3
r
(
2
−
2
r
)
(
r
+
1
)
(
r
+
2
)
is equal to?
Report Question
0%
1
2
−
3
100
100
(
101
)
0%
3
2
−
3
101
101
(
102
)
0%
3
2
−
3
100
100
(
101
)
0%
None of these
Explanation
S
=
100
∑
r
=
2
3
r
(
2
−
2
r
)
r
(
r
+
1
)
(
r
+
2
)
S
=
100
∑
r
=
2
3
r
r
(
r
+
2
)
−
3
r
+
1
(
r
+
1
)
(
r
+
2
)
=
3
2
2
(
3
)
−
3
3
3
×
4
+
3
3
3
×
4
−
3
4
4
×
5
+
3
4
4
×
5
+
.
.
.
+
(
3
100
100
×
101
−
3
101
101
×
102
)
=
3
2
2
(
3
)
−
3
101
101
×
102
S
=
3
2
−
3
101
101
×
102
The sum of
n
terms of the series
1
3
+
3
3
+
5
3
+
…
is:
Report Question
0%
3
n
2
(
2
n
2
−
1
)
0%
n
2
(
2
n
2
−
1
)
0%
n
2
(
n
2
−
1
)
0%
n
2
(
2
n
2
−
5
)
If
f
(
x
)
=
Π
3
i
=
1
(
x
−
a
i
)
+
3
∑
i
=
1
a
i
−
3
x
where
a
i
<
a
i
+
1
for
i
=
1
,
2
,
then
f
(
x
)
=
0
has:
Report Question
0%
Only one distinct real root
0%
Exactly two distinct real root
0%
Exactly
3
distinct real roots
0%
3
equal real roots
The sum of
7
2
×
3
(
1
3
)
+
9
3
×
4
(
1
3
)
2
+
11
4
×
5
(
1
3
)
3
+
upto 10 terms is equal to
Report Question
0%
1
2
−
1
12
×
3
10
0%
1
3
−
1
12
×
3
10
0%
1
2
−
1
10
×
3
10
0%
N
o
n
e
o
f
t
h
e
s
e
The sum of the series
(
2
)
2
+
2
(
4
)
2
+
3
(
6
)
2
+
.
.
.
.
upto
10
terms is
Report Question
0%
12100
0%
11300
0%
11200
0%
12300
Explanation
(
2
)
2
+
2
(
4
)
2
+
3
(
6
)
2
+
…
.
.
u
p
t
o
10
t
e
r
m
T
m
=
n
(
2
n
)
2
⇒
4
n
3
10
∑
n
=
0
T
n
=
4
10
∑
n
=
0
n
3
=
4
(
n
)
2
(
n
+
1
)
2
4
n
=
10
→=
100
×
121
=
12100
The value of
10
∑
x
=
1
r
=
x
−
1
∑
r
=
0
(
2
x
−
2
r
)
is
Report Question
0%
16392
0%
16398
0%
15462
0%
15468
Explanation
10
∑
x
=
1
x
−
1
∑
r
=
0
(
2
x
−
2
r
)
Now,
x
−
1
∑
r
=
0
(
2
x
−
2
r
)
=
(
2
x
−
2
1
)
+
(
2
x
−
2
2
)
+
.
.
.
.
.
.
+
(
2
x
−
2
x
−
1
)
=
(
2
x
+
2
x
+
2
x
+
.
.
.
.
.
.
.
.
+
x
t
e
r
m
s
)
−
(
2
0
+
2
1
+
2
2
+
.
.
.
.
.
.
+
2
x
−
1
)
=
x
.2
x
−
[
2
0
(
2
x
−
1
)
2
−
1
]
=
x
.2
x
−
(
2
x
−
1
)
=
x
.2
x
−
2
x
+
1
Now,
10
∑
x
=
1
x
−
1
∑
r
=
0
(
2
c
−
2
r
)
=
10
∑
x
=
1
x
.2
x
+
10
∑
x
=
1
1
=
10
∑
x
=
1
x
.2
x
−
[
2
1
+
2
2
+
2
3
+
.
.
.
.
.
.
.
.
+
2
10
]
+
10
=
10
∑
x
=
1
x
.2
x
−
[
2
(
2
10
−
1
)
2
−
1
]
+
10
=
10
∑
x
=
1
x
.2
x
−
2
11
+
12
=
(
1.2
+
2.2
2
+
3.2
3
+
.
.
.
.
.
.
+
10.2
10
)
−
2
11
+
12
=
18734
−
2048
+
12
=
16398
Ans
Observe the pattern carefully
11
×
11
=
121
111
×
111
=
12321
1111
×
1111
=
?
Report Question
0%
12345321
0%
123421
0%
1234321
0%
12468
Explanation
11
×
11
=
121
111
×
111
=
12321
1111
×
1111
=
1234321
Sum of the series
(
1
×
2015
)
+
(
2
×
2014
)
+
(
3
×
2013
)
.
.
.
.
.
.
+
(
2015
×
1
)
is equal to-
Report Question
0%
336
×
2015
×
2016
0%
336
×
2015
×
2017
0%
336
×
2016
×
201
0%
N
o
n
e
It is known that
∞
∑
r
=
1
1
(
2
r
−
1
)
2
=
π
2
8
,
then
∞
∑
r
=
1
1
r
2
is equal to
Report Question
0%
π
2
24
0%
π
2
3
0%
π
2
6
0%
none of these
Explanation
Given :
∞
∑
r
=
1
1
(
2
r
−
1
)
2
=
π
2
8
,
∞
∑
r
=
1
1
r
2
=
?
?
start putting 'r' value
1
1
2
+
1
3
2
+
1
5
2
+
1
7
2
−
−
−
−
−
−
=
π
2
8
let say
∞
∑
r
=
1
1
r
2
=
k
1
1
2
+
1
2
2
+
1
3
2
+
1
4
2
+
−
−
−
−
−
1
r
2
=
k
[
1
1
2
+
1
3
2
+
1
5
2
+
−
−
−
−
−
1
(
2
r
−
1
)
2
]
+
1
2
2
+
1
4
2
+
−
−
−
−
=
k
π
2
8
+
1
2
2
[
1
1
2
+
1
2
2
+
1
3
2
−
−
−
−
]
=
k
π
2
8
+
1
4
(
k
)
=
k
π
2
8
=
k
−
k
4
=
3
k
4
so,
k
=
π
2
6
Answer option- C
Find the sum of the infinite series
1
9
+
1
81
+
1
729
+
.
.
.
.
∞
Report Question
0%
1
8
0%
1
4
0%
1
5
0%
2
3
The sum
3
1.2
,
3
1.2
,
1
2
,
4
2.3
(
1
2
)
2
+
5
3.4
(
1
2
)
2
Report Question
0%
1
−
1
(
n
+
1
)
2
n
0%
1
−
1
n
.2
n
−
1
0%
1
=
1
(
n
+
1
)
2
n
0%
1
(
n
−
1
)
2
n
−
1
Explanation
t
n
=
n
+
2
n
(
n
+
1
)
⋅
(
1
2
)
n
=
2
(
n
+
1
)
−
n
n
(
n
+
1
)
⋅
(
1
2
)
n
t
n
=
1
n
(
1
2
)
n
−
1
−
1
n
+
1
⋅
(
1
2
)
n
S
n
=
n
∑
n
=
1
t
n
=
{
1
1
(
1
2
)
0
−
1
2
(
1
2
)
1
}
+
{
1
2
(
1
2
)
1
−
1
3
(
1
2
)
2
}
+
.
.
.
+
=
{
1
n
(
1
2
)
n
−
1
−
1
n
+
1
(
1
2
)
n
}
S
n
=
1
−
1
(
n
+
1
)
2
n
If f(x)=
x
+
1
2
x
+
1
1
2
x
+
.
.
.
∞
;
then the value of f (2011). f'(2011) is :
Report Question
0%
0
0%
1
0%
2011
0%
2010
1
+
n
2
+
n
(
n
−
1
)
2.4
+
n
(
n
−
1
)
(
n
−
2
)
2.4
.6
+
…
…
…
∞
=
?
Report Question
0%
(
2
3
)
n
0%
(
3
2
)
n
0%
(
3
4
)
n
0%
(
4
3
)
n
Find the missing number, if same rule is followed in all the three figures.
Report Question
0%
12
0%
16
0%
14
0%
9
Explanation
REF.Image.
Given
If we consider down two number of fig(i)
then their product is
⇒
4
×
16
=
64
Now if we take square root then =
√
64
=
8
→
upper are we are getting.
similarly from fig (iii)
down number product
⇒
12
×
27
=
324
take square roots
=
√
324
=
18
→
upper number.
So, similarly from fig(ii)
Down product =
18
×
8
=
144
so square root =
√
144
=
12
→
option A
Choose the
CORRECT
options:-
Report Question
0%
360
∑
k
=
1
(
1
k
√
k
+
1
+
(
k
+
1
)
√
k
)
is the ratio of two relative prime positive integers
m
nad
n
. The value of
(
m
+
n
)
is equal to
37
.
0%
If
5
π
2
<
x
<
3
π
, then the value of the expression
√
1
−
sin
x
+
√
1
+
sin
x
√
1
−
sin
x
−
√
1
+
sin
x
is
−
tan
x
2
.
0%
The exact value of
96
sin
80
∘
sin
65
∘
sin
35
∘
sin
20
∘
+
sin
50
∘
+
sin
110
∘
is equal to
24
.
0%
The sum
∞
∑
n
=
1
(
n
n
4
+
4
)
is equal to
3
/
16
.
n
∑
r
=
1
sin
−
1
(
√
r
−
√
r
−
1
√
r
(
r
+
1
)
)
is equal to
Report Question
0%
tan
−
1
(
√
n
)
−
π
4
0%
tan
−
1
(
√
n
+
1
)
−
π
4
0%
tan
−
1
(
√
n
)
0%
tan
−
1
(
√
n
+
1
)
Direction: In a given question, one number is missing in the series. you have to understand the pattern of the series and insert the number
JN 28 27 GP
CE 12 45 TU
LR ? ? MS
Report Question
0%
34
,
36
0%
35
,
35
0%
30
,
32
0%
30
,
41
Let
A
be the sum of the first
20
terms and
B
be the sum of the first
40
terms of the series
1
+
2.2
2
+
3
2
+
2.4
2
+
5
2
+
2.6
2
+......... Find the value of
A
.
Report Question
0%
496
0%
232
0%
248
0%
464
Explanation
Sum of first
20
terms -
⇒
(
1
2
+
3
2
+
_
_
_
_
+
19
2
)
+
2
(
2
2
+
4
2
+
_
_
_
_
+
20
2
)
⇒
A
=
(
1
2
+
2
2
+
3
2
_
_
_
_
+
19
2
+
20
2
)
+
(
2
2
+
4
2
+
_
_
_
_
+
20
2
)
⇒
A
=
20
×
41
×
21
6
+
1
2
(
1
2
+
2
2
+
_
_
_
_
+
10
2
)
⇒
A
=
2870
+
4
×
10
×
11
×
21
6
⇒
A
=
2870
+
1540
⇒
A
=
4410.
Hence, the answer is
4410.
If
1
2
+
2
2
+
3
2
+
+
(
2003
)
2
=
(
2003
)
(
4007
)
(
334
)
and
(
1
)
(
2003
)
+
(
2
)
(
2002
)
=
(
2003
)
(
334
)
(
x
)
then
x
equals
Report Question
0%
2005
0%
2004
0%
2003
0%
2001
The number of the three digits numbers having only two consecutive digits identical is:
Report Question
0%
153
0%
160
0%
180
0%
161
The sum of the first n terms of the series
1
2
+
3
4
+
7
8
+
15
16
+
is equal to:
Report Question
0%
2
−
n
−
n
−
1
0%
1
−
2
−
n
0%
n
+
2
−
n
−
1
0%
2
−
n
−
1
1
+
(
1
+
2
)
+
(
1
+
2
+
2
2
)
+
(
1
+
2
+
2
2
+
2
3
)
=
.
.
.
.
upto
n
terms
=
______
Report Question
0%
2
n
+
2
−
n
−
4
0%
2
(
2
n
−
1
)
−
n
0%
2
n
+
1
−
n
0%
2
2
n
+
1
−
1
If
24
+
37
=
7
,
12
+
18
=
3
, then 54+21=?
Report Question
0%
11
0%
9
0%
12
0%
3
In the following number series,one number is wrong.find out the ?
1,2,8,33,149,765,4626
Report Question
0%
33
0%
8
0%
149
0%
0
Explanation
1,2,8,33,149,765,4626
Carefully analyzing the series, we can see the following pattern
1
×
1
+
(
1
)
2
=
2
2
×
2
+
(
2
)
2
=
8
8
×
3
+
(
3
)
2
=
33
33
×
4
+
(
4
)
2
=
148
148
×
5
+
(
5
)
2
=
765
765
×
6
+
(
6
)
2
=
4626
So, we can clearly see that 149 is the wrong entry in the series.
If
′
−
′
denotes
′
÷
′
,
′
÷
′
denotes
′
×
′
,
′
+
′
denotes' - 'and
′
×
′
denotes
′
+
′
, then find the value of
116
+
9
÷
52
−
4
×
5
.
Report Question
0%
16
0%
8
0%
9
0%
4
Explanation
−
denoted
÷
÷
denoted
×
+
denoted
−
×
denoted
+
B
O
D
M
A
S
116
+
9
÷
52
−
4
×
5
116
−
9
×
52
÷
4
+
5
116
−
9
×
13
+
5
116
−
117
+
5
116
−
112
=
4
1
+
1
2
(
1
+
2
)
+
1
3
(
1
+
2
+
3
)
+
1
4
(
1
+
2
+
3
+
4
)
+
.
.
.
.
.
upto
20
terms is
Report Question
0%
110
0%
111
0%
115
0%
116
Explanation
Given
S
=
1
+
1
2
(
1
+
2
)
+
1
3
(
1
+
2
+
3
)
+
.
.
.
.
.
+
1
20
(
1
+
2
+
3
+
.
.
.
.
+
20
)
=
20
∑
n
=
1
1
n
∑
n
20
∑
n
=
1
1
n
.
n
(
n
+
1
)
2
=
1
2
20
∑
n
=
1
(
n
+
1
)
1
2
[
n
(
n
+
1
)
2
+
n
]
20
0
=
1
2
[
20
(
21
)
2
+
20
]
=
5
(
21
)
+
10
=
115
Find the missing term in the series given below.
12
,
13
,
18
,
19
,
24
,
25
?
Report Question
0%
27
0%
30
0%
29
0%
26
Explanation
12
,
13
,
18
,
19
,
24
,
25
difference between
12
,
13
=
1
difference between
13
,
18
=
5
Similarly
difference between
18
,
19
=
1
difference between
19
,
24
=
5
similarly
difference between
24
,
25
=
1
so to get next number of
25
add
5
i.e.,
25
+
5
=
30
Let a sequence whose
n
t
h
term is
a
n
be define as
a
1
=
1
2
and
(
n
−
1
)
a
n
−
1
=
(
n
+
1
)
a
n
for
n
≥
2
; then
Report Question
0%
a
n
=
1
n
(
n
+
1
)
0%
S
n
=
1
−
1
n
+
1
0%
lim
n
→
∞
S
n
=
1
0%
S
n
=
1
n
+
1
Explanation
a
1
=
1
2
(
n
+
1
)
a
n
=
(
n
−
1
)
a
n
−
1
,
n
≥
2
For,
n
=
2
,
3
a
2
=
a
1
n
=
3
,
4
a
3
=
2
a
2
n
=
4
,
5
a
4
=
3
a
3
.................................
n
=
n
,
(
n
+
1
)
a
n
=
(
n
−
1
)
a
n
−
1
Multiplying the above, we get
3
a
2
×
4
a
3
×
5
a
4
.
.
.
.
.
.
.
.
.
.
.
×
(
n
+
1
)
a
n
=
a
1
×
2
a
2
×
3
a
3
×
4
a
4
×
.
.
.
.
.
.
.
.
.
.
.
.
.
.
×
(
n
−
1
)
a
n
−
1
⇒
3
a
2
×
4
a
3
×
5
a
4
.
.
.
.
.
.
.
.
.
.
.
×
(
n
+
1
)
a
n
=
1
2
×
2
a
2
×
3
a
3
×
4
a
4
×
.
.
.
.
.
.
.
.
.
.
.
.
.
.
×
(
n
−
1
)
a
n
−
1
⇒
3
a
2
×
4
a
3
×
5
a
4
.
.
.
.
.
.
.
.
.
.
.
×
(
n
)
a
n
−
1
(
n
+
1
)
a
n
=
1
2
×
2
a
2
×
3
a
3
×
4
a
4
×
.
.
.
.
.
.
.
.
.
.
.
.
.
.
×
(
n
−
1
)
a
n
−
1
⇒
n
(
n
+
1
)
a
n
=
1
⇒
a
n
=
1
n
(
n
+
1
)
=
1
n
−
1
n
+
1
∴
\Rightarrow S_{n}=1-\cfrac{1}{n+1}
Now,
\lim_{n\rightarrow \infty}S_{n}=\lim_{n\rightarrow \infty}1-\cfrac{1}{n+1}=1-\lim_{n\rightarrow \infty}\cfrac{1}{n+1}=1
The sum of the series 1.2.3+2.3.4+...3.4.5.+...to n terms is
Report Question
0%
n(n+1)(n+2)
0%
(n+1)(n+2)(n+3)
0%
\frac { 1 }{ 4 } n(n+1)(n+2)(n+3)
0%
\frac { 1 }{ 4 } (n+1)(n+2)(n+3)
The sum of first
9
terms of the series
\dfrac { { 1 }^{ 3 } }{ 1 } +\dfrac { { 1 }^{ 3 }+{ 2 }^{ 3 } }{ 1+3 } +\dfrac { { 1 }^{ 3 }+{ 2 }^{ 3 }+{ 3 }^{ 3 } }{ 1+3+5 } +\dots
Report Question
0%
96
0%
142
0%
192
0%
71
If
\left( { 1-y } \right) ^{ 35 }\left( 1+y \right) ^{ 45 }={ A }_{ 0 }+{ A }_{ 1 }Y+{ A }_{ 2 }{ Y }^{ 2 }+{ A }_{ 3 }{ Y }^{ 3 }+.....+{ A }_{ 80 }{ Y }^{ 80 },
then
Report Question
0%
\frac { { A }_{ 2 } }{ A_{ 1 } } <2
0%
{ A }_{ 1 }={ A }_{ 2 }
0%
\frac { { A }_{ 2 } }{ { A }_{ 1 } } <1
0%
1<\frac { { A }_{ 2 } }{ { A }_{ 1 } } <2
If
\dfrac{1}{1^{2}}+\dfrac{1}{2^{2}}+\dfrac{1}{3^{2}}+........\infty =\dfrac{\pi^{2}}{6}
then
\dfrac{1}{1^{2}}+\dfrac{1}{3^{2}}+\dfrac{1}{5^{2}}+......\infty
Report Question
0%
\dfrac{\pi^{2}}{8}
0%
\dfrac{\pi^{2}}{12}
0%
\dfrac{\pi^{2}}{3}
0%
\dfrac{\pi^{2}}{2}
Sum of the series
1+2.2+3.2^{2}+...+100.2^{10}=
Report Question
0%
100.2^{100}+1
0%
99.2^{100}+1
0%
99.2^{100}-1
0%
100.2^{100}-1
Sum of series
\displaystyle \sum^{n}_{r=1}(r^{2}+1)_{r!}
is
Report Question
0%
(n+1)!
0%
(n+2)!-1
0%
n(n+2)!
0%
(n+1)!-1
\dfrac {1}{1.4}+\dfrac {1}{4.7}+\dfrac {1}{7.10}+...+\dfrac {1}{(3n-5)(3n-2)}
Report Question
0%
\dfrac {n-1}{3n+2}
0%
\dfrac {n-1}{3n-2}
0%
\dfrac {n-1}{3n-1}
0%
\dfrac {n-1}{3n-4}
If
x=\sqrt 1+\dfrac{1}{{1}^{2}}+\dfrac{1}{{2}^{2}}+\sqrt 1+\dfrac{1}{{2}^{2}}+\dfrac{1}{{3}^{2}}+\sqrt 1+\dfrac{1}{{2019}^{2}}+\dfrac{1}{{2020}^{2}}
then
4040(2020-x)
Report Question
0%
1
0%
2
0%
-1
0%
-2
The sum of the series
1+\frac { 1.3 }{ 6 } +\frac { 1.3.5 }{ 6.8 } +...\infty
is
Report Question
0%
4
0%
0
0%
\infty
0%
{ (4) }^{ \frac { 2 }{ 3 } }
The sum of the infinite series
\cot { ^{ 1 } } \left( \dfrac { 7 }{ 4 } \right) +\cot { ^{ -1 }\left( \dfrac { 19 }{ 4 } \right) } +\cot { ^{ -1 }\left( \dfrac { 39 }{ 4 } \right) } +\cot { ^{ -1 }\left( \dfrac { 67 }{ 4 } \right) } +........\infty
is:
Report Question
0%
\\ \dfrac { \pi }{ 4 } -\cot { ^{ -1 } } (3)
0%
\\ \dfrac { \pi }{ 4 } -\tan { ^{ -1 } } (3)
0%
\\ \dfrac { \pi }{ 4 } +\cot { ^{ -1 } } (3)
0%
\\ \dfrac { \pi }{ 4 } +\tan { ^{ -1 } } (3)
40280625, 732375, 16275, 465, 18.6, 1.24,?
Report Question
0%
0.248
0%
0.336
0%
0.424
0%
0.512
0%
0.639
Explanation
40280625, 732375,16275,465,18.6,1.24
\cfrac{40280625}{732375}=55
\cfrac{732375}{16275}=45
\cfrac{16275}{465}=35
\cfrac{465}{18.6}=25
\cfrac{18.6}{1.24}=15
\cfrac{1.24}{x}=5
x=0.248
Sum of the series
S = 1^{2}-2^{2}+3^{2}-4^{2}+......-2002^{2}+2003^{2}
is
Report Question
0%
2007006
0%
1005004
0%
2000506
0%
none of these
In the sum
3+33+333+3333+.........2015
terms the number formed by taking the last four digits in that order is
Report Question
0%
6365
0%
6255
0%
6465
0%
6565
Sum of the series
\dfrac{1^2}{1!}+\dfrac{2^2}{2!}+\dfrac{3^2}{3!}+\dfrac{4^2}{4!}+.....
(infinite terms) is?
Report Question
0%
e
0%
2e
0%
e^2
0%
\infty
Explanation
\begin{aligned} &S=\frac{1^{2}}{1 !}+\frac{2^{2}}{2 !}+\frac{3^{2}}{3 !}+\frac{4^{2}}{4 b}+\\ &\text {(infinite terems)} is \\ &S=\sum_{n=1}^{\infty} n^{2}\\ &n^{2}=a_{0}+a_{1} n+a_{2} n(n-1)\\ &n=0 \quad 0=a_{0}\\ &n=1 \quad 1=0+a, \Rightarrow a_{1}=1\\ &n=2 \quad 4=0+2+2 a_{2} \Rightarrow a_{2}=1\\ &S=\sum_{n=1}^{\infty}\left(a_{0}+a_{1} n+a_{2} n(n-1)\right) \end{aligned}
S=\sum_{n=1}^{\infty}\left(\frac{0+n+n(n-1)}{n_{0}}\right)
S=\sum_{n=1}^{\infty} \frac{n}{n_{b}}+\sum_{n=1}^{\infty} \frac{n(n-1)}{n_{0}^{1}}
S=e+e=2 e
e=1+\frac{1+\frac{1}{2 !}+\frac{1}{3 !}+\frac{1}{4 !}+\cdots} \Rightarrow \sum_{n=1}^{\infty} \frac{1}{(n-1) !}
e=\sum_{n=2(n-2)\}}=1+1+\frac{1}{2 !}+\frac{1}{3 !} \cdots \cdot \cdots
Option
B
is correct
If
a_{k}=\dfrac{1}{K(K+1)}
for
K=1, 2, 3,.....n,
then
\left(\sum_{k=1}^{n}{a_{k}}\right)^{2}=
Report Question
0%
\dfrac{n}{n+1}
0%
\dfrac{n^{2}}{(n+1)^{2}}
0%
\dfrac{n^{4}}{(n+1)^{4}}
0%
\dfrac{n^{6}}{(n+1)^{6}}
Explanation
\begin{array}{l} { a_{ k } }=\dfrac { 1 }{ { k\left( { k+1 } \right) } } =\dfrac { 1 }{ k } -\dfrac { 1 }{ { k+1 } } \\ { \left( { \sum _{ k=1 }^{ n }{ { a_{ x } } } } \right) ^{ 2 } } \\ { \left( { \sum _{ k=1 }^{ n }{ \left( { \dfrac { 1 }{ k } -\dfrac { 1 }{ { k+1 } } } \right) } } \right) ^{ 2 } } \\ =\left( { 1-\dfrac { 1 }{ 2 } } \right) +\left( { \dfrac { 1 }{ 2 } -\dfrac { 1 }{ 3 } } \right) \, \, \, \, \, \left( { \dfrac { 1 }{ n } \dfrac { { -1 } }{ { n+1 } } } \right) \\ ={ \left( { 1-\dfrac { 1 }{ { n+1 } } } \right) ^{ 2 } } \\ ={ \left( { \dfrac { { n+1-1 } }{ { n+1 } } } \right) ^{ 2 } } \\ =\dfrac { { { n^{ 2 } } } }{ { { { \left( { n+1 } \right) }^{ 2 } } } } \end{array}
Hence, this is the answer.
\dfrac { { C }_{ 1 } }{ { C }_{ 0 } } +2\dfrac { { C }_{ 2 } }{ { C }_{ 1 } } +3\frac { { C }_{ 3 } }{ { C }_{ 2 } } +......+\dfrac { n{ C }_{ n } }{ { C }_{ n-1 } }
equals
Report Question
0%
{ n2 }^{ n-1 }
0%
\dfrac { { n2 }^{ n-1 } }{ { 2 }^{ n }-1 }
0%
\dfrac { n(n+1) }{ 2 }
0%
\dfrac { (n+1)(n+2) }{ 2 }
0:0:2
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50
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48
49
50
0
Answered
1
Not Answered
49
Not Visited
Correct : 0
Incorrect : 0
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Practice Class 11 Engineering Maths Quiz Questions and Answers
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