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KEAM 2016 Math Paper
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© examsnet.com
Question : 22
Total: 120
Sum of the series 1(1) + 2 ( 1 + 3) + 3 (1 + 3 +5) + 4 (1 + 3 + 5 + 7) + ….+ 10 (1 + 3 + 5 + 7+ …+ 19) is equal to
385
1025
1125
2025
3025
Validate
Solution:
1
(
1
)
+
2
(
1
+
3
)
+
3
(
1
+
3
+
5
)
+
4
(
1
+
3
+
5
+
7
)
+
.
.
.
+
10
(
1
+
3
+
5
+
7
+
.
.
.
+
19
)
=
1
×
1
+
2
×
2
2
+
3
×
3
2
+
4
×
4
2
+
.
.
.
+
10
×
10
2
=
1
3
+
2
3
+
3
3
+
4
3
+
.
.
.
+
10
3
=
[
10
(
10
+
1
)
2
]
2
[ ∵ sum of cubes of
n
natural numbers
=
n
(
n
+
1
)
2
2
]
=
[
10
×
11
2
]
2
=
(
55
)
2
=
3025
© examsnet.com
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