Tardigrade
Tardigrade - CET NEET JEE Exam App
Exams
Login
Signup
Tardigrade
Question
Mathematics
The sum of the series 1+2 × 3+4+5 × 6+7+8 × 9+ ... up to 50 terms is
Q. The sum of the series
1
+
2
×
3
+
4
+
5
×
6
+
7
+
8
×
9
+
... up to
50
terms is
2508
220
Sequences and Series
Report Error
A
48775
B
47825
C
49675
D
50000
Solution:
S
50
=
r
=
1
∑
25
[(
3
r
−
2
)
+
(
3
r
−
1
)
(
3
r
)]
=
r
=
1
∑
25
(
9
r
2
−
2
)
=
6
9
(
25
)
(
26
)
(
51
)
−
50
=
49675