Express 5055 as a product of prime numbers
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Answer:
7
th
term is 16
C
sum of first 10 terms is
4
505
Given series is,
S=1+
(1+3)
(1+2)
2
+
(1+3+5)
(1+2+3)
2
+
(1+3+5+7)
(1+2+3+4)
2
+…
The r
th
term is given by,
T
r
=
{1+3+5+...+(2r−1)}
(1+2+3+...+r)
2
=
r
2
⋅2
2
{r(r+1)}
2
=
4
r
2
+2r+1
Hence, T
7
=
4
7
2
+2×7+1
=
4
64
=16
Now, S
∞
=
r=1
∑
∞
T
r
∴S
10
=
r=1
∑
10
T
r
=
4
1
(
r=1
∑
10
r
2
+2
r=1
∑
10
r+
r=1
∑
10
1)
=
4
1
(
6
10(10+1)(20+1)
+2⋅
2
10(10+1)
+10)
=
4
1
(
6
10⋅11⋅21
+10⋅11+10)
=
4
505
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