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Let’s check out your problem:
(
8
8
8
) If the number of terms in
(
x
+
1
+
1
x
)
n
(
n
∈
I
+
)
\left(x+1+\frac{1}{x}\right)^{n}\left(n \in I^{+}\right)
(
x
+
1
+
x
1
)
n
(
n
∈
I
+
)
is
401
401
401
, then
n
n
n
is greater than
\newline
(a)
201
201
201
\newline
(b)
200
200
200
\newline
(c)
199
199
199
\newline
(d) None of these
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Math Problems
Algebra 2
Sum of finite series not start from 1
Full solution
Q.
(
8
8
8
) If the number of terms in
(
x
+
1
+
1
x
)
n
(
n
∈
I
+
)
\left(x+1+\frac{1}{x}\right)^{n}\left(n \in I^{+}\right)
(
x
+
1
+
x
1
)
n
(
n
∈
I
+
)
is
401
401
401
, then
n
n
n
is greater than
\newline
(a)
201
201
201
\newline
(b)
200
200
200
\newline
(c)
199
199
199
\newline
(d) None of these
General Term Expansion:
The general term in the expansion of
(
x
+
1
+
1
x
)
n
(x+1+\frac{1}{x})^n
(
x
+
1
+
x
1
)
n
is
T
k
+
1
=
C
(
n
,
k
)
⋅
x
n
−
k
⋅
(
1
x
)
k
T_{k+1} = C(n, k) \cdot x^{n-k} \cdot \left(\frac{1}{x}\right)^k
T
k
+
1
=
C
(
n
,
k
)
⋅
x
n
−
k
⋅
(
x
1
)
k
.
Number of Terms:
The number of terms in the expansion is
n
+
1
n+1
n
+
1
because the powers of
x
x
x
range from
x
n
x^n
x
n
to
x
−
n
x^{-n}
x
−
n
.
Find
n
n
n
:
To find
n
n
n
when there are
401
401
401
terms, we set
n
+
1
=
401
n+1 = 401
n
+
1
=
401
.
Solve for n:
Solve for n:
n
=
401
−
1
n = 401 - 1
n
=
401
−
1
.
Compare to Options:
n
=
400
n = 400
n
=
400
.
Compare to Options:
n
=
400
n = 400
n
=
400
.Compare
n
=
400
n = 400
n
=
400
to the options given: (a)
201
201
201
, (b)
200
200
200
, (c)
199
199
199
, (d) None of these.
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Choices:Choices:
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∑
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\text{[A]arithmetic}
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[B]geometric
\text{[B]geometric}
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[C]both
\text{[C]both}
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[D]neither
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Question
Find the first three partial sums of the series.
\newline
1
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21
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26
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21
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Write your answers as integers or fractions in simplest form.
\newline
S
1
=
S_1 =
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1
=
____
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S
2
=
S_2 =
S
2
=
____
\newline
S
3
=
S_3 =
S
3
=
____
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Find the third partial sum of the series.
\newline
3
+
9
+
15
+
21
+
27
+
33
+
⋯
3 + 9 + 15 + 21 + 27 + 33 + \cdots
3
+
9
+
15
+
21
+
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+
33
+
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\newline
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\newline
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Find the first three partial sums of the series.
\newline
1
+
7
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13
+
19
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25
+
31
+
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1 + 7 + 13 + 19 + 25 + 31 + \cdots
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+
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+
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+
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\newline
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\newline
S
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=
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1
=
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\newline
S
2
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=
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\newline
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Question
Does the infinite geometric series converge or diverge?
\newline
1
+
3
4
+
9
16
+
27
64
+
⋯
1 + \frac{3}{4} + \frac{9}{16} + \frac{27}{64} + \cdots
1
+
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\newline
Choices:
\newline
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[A]converge
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[B]diverge
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