Find sum of the series (1+2)+(1+2+2²)+(1+2+2²+2³)+.... upto n terms.
Answers
Answer:
The n
th
term of the numerator =n(n+1)
2
=n
2
+2n
2
+n
And n
th
term of the denominator =n
2
(n+1)=n
3
+n
2
∴
1
2
×2+2
2
×3+...+n
2
×(n+1)
1×2
2
+2×3
2
+...+n×(n+1)
2
=
∑
k−1
n
a
k
∑
k−1
n
a
k
=
∑
k−1
n
(k
3
+k
2
)
∑
k−1
n
(k
3
+k
2
+k)
...(1)
Here,
k−1
∑
n
(k
3
+2k
2
+k)
=
4
n
2
(n+1)
2
+
6
2n(n+1)(2n+1)
+
2
n(n+1)
=
2
n(n+1)
[
2
n(n+1)
+
3
2
(2n+1)+1]
=
2
n(n+1)
[
6
3n
2
+3n+8n+4+6
]
=
12
n(n+1)
[3n
2
+11n+10]
=
12
n(n+1)
[3n
2
+6n+5n+10]
=
12
n(n+1)
[3n(n+2)+5(n+2)]
=
12
n(n+1)(n+2)(3n+5)
...(2)
Also,
k−1
∑
n
(k
3
+k
2
)=
4
n
2
(n+1)
2
+
6
n(n+1)(2n+1)
=
2
n(n+1)
[
2
n(n+1)
+
3
2n+1
]
=
2
n(n+1)
[
6
3n
2
+3n+4n+2
]
=
12
n(n+1)
[3n
2
+7n+2]
=
12
n(n+1)
[3n
2
+6n+n+2]
=
12
n(n+1)
[3n(n+2)+1(n+2)]
=
12
n(n+1)(n+2)(3n+1)
...(3)
From (1), (2) and (3) we obtain
1
2
×2+2
2
×3+...+n
2
×(n+1)
1×2
2
+2×3
2
+...+n×(n+1)
2
=
12
n(n+1)(n+2)(3n+1)
12
n(n+1)(n+2)(3n+5)
=
n(n+1)(n+2)(3n+1)
n(n+1)(n+2)(3n+5)
=
3n+1
3n+5
Thus, the given result is proved.