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TG EAMCET 2 May 2025 Shift 1 Paper
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© examsiri.com
Question :
3 of 160
Marks:
+1
,
-0
1
+
(
1
+
3
)
+
(
1
+
3
+
5
)
+
(
1
+
3
+
5
+
7
)
+
.
.
.
to 10 terms
=
385
285
506
406
Validate
Solution:
Given series
1
+
(
1
+
3
)
+
(
1
+
3
+
5
)
+
(
1
+
3
+
5
+
7
)
+
.
.
.
to 10 terms.
The
n
th term of the series is the sum of first
n
odd numbers.
So,
T
n
=
1
+
3
+
5
+
.
.
.
+
(
2
n
−
1
)
=
n
2
∴
‌
T
1
=
1
2
=
1
‌
T
2
=
2
2
=
4
‌
T
3
=
3
2
=
9
T
10
=
10
2
=
100
Thus,
S
10
=
10
∑
n
=
1
n
2
We know that the sum of the first
k
squares is
k
∑
n
=
1
n
2
=
‌
k
(
k
+
1
)
(
2
k
+
1
)
6
For
k
=
10
S
10
‌
=
‌
10
(
10
+
1
)
(
2
×
10
+
1
)
6
‌
=
‌
10
×
11
×
21
6
=
‌
2310
6
=
385
∴
S
10
‌
=
385
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