the sum upto infinity of the series (1+2^-2)+(2^-1+2^-4)+(2^-2+2^-6)+...... is
Answers
= 1(2) + 2(2²) + 3(2³) +... + 100(2^100)
This is an arithmetico geometric series , with A.P : 1,2,3,...,100
G.P : 2,2²,2³,....,2^100
Now,
S = 1(2) + 2(2²) + 3(2³) + ..... + 100(2^100) .........(i)
.°. 2S = 1(2²) + 2(2³) + 3(2⁴) + .... + 99(2^100) + 100(2^101) ......(ii)
Subtracting equation (ii) from (i) we get
-S = 2 + 2²(2-1)+2³(3-2)+....+2^100(100-99) - 100(2^101)
= (2 + 2² + 2³ + .... + 2^100) - 100(2^101)
.°. -S = 2[(2^100) - 1] - 100(2^100)
100-1
.°. S = 100(2^100) - 2[(2^100) - 1]
99
Answer:
Step-by-step explanation:We will use the following known sums (each of which can be proven via induction):
n
∑
i
=
1
i
=
n
(
n
+
1
)
2
n
∑
i
=
1
i
2
=
n
(
n
+
1
)
(
2
n
+
1
)
6
n
∑
i
=
1
(
i
3
)
=
n
2
(
n
+
1
)
2
4
With those:
(
1
2
)
+
(
1
2
+
2
2
)
+
...
+
(
1
2
+
2
2
+
...
+
n
2
)
=
n
∑
i
=
1
(
1
2
+
2
2
+
...
+
i
2
)
=
n
∑
i
=
1
i
∑
j
=
1
j
2
=
n
∑
i
=
1
i
(
i
+
1
)
(
2
i
+
1
)
6
=
n
∑
i
=
1
2
i
3
+
3
i
2
+
i
6
=
n
∑
i
=
1
1
3
i
3
+
n
∑
j
=
1
1
2
j
2
+
n
∑
k
=
1
1
6
k
=
1
3
n
∑
i
=
1
i
3
+
1
2
n
∑
j
=
1
j
2
+
1
6
n
∑
k
=
1
k
=
1
3
(
n
2
(
n
+
1
)
2
4
)
+
1
2
(
n
(
n
+
1
)
(
2
n
+
1
)
6
)
+
1
6
(
n
(
n
+
1
)
2
)
=
n
2
(
n
+
1
)
2
12
+
n
(
n
+
1
)
(
2
n
+
1
)
12
+
n
(
n
+
1
)
12
=
n
(
n
+
1
)
[
n
(
n
+
1
)
+
(
2
n
+
1
)
+
1
]
12
=
n
(
n
+
1
)
(
n
2
+
3
n
+
2
)
12
=
n
(
n
+
1
)
2
(
n
+
2
)
12