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Byju's Answer
Standard XII
Mathematics
General Term of Binomial Expansion
The number of...
Question
The number of integral terms in the expansion of
(
7
1
/
3
+
11
1
/
9
)
6561
is
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Solution
T
r
+
1
=
6561
C
r
(
7
1
3
)
6561
−
r
(
11
1
9
)
r
=
6561
C
r
7
(
2187
−
r
3
)
11
(
r
9
)
;
r
must be multiple of
9.
r
=
0
,
9
,
18
,
⋯
,
6561
⇒
6561
=
0
+
(
k
−
1
)
×
9
⇒
k
=
730
Suggest Corrections
4
Similar questions
Q.
In the expansion of
(
7
1
3
+
11
1
9
)
6561
,
Q.
In the expansion of
(
7
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6561
Q.
In the expansion of
(
7
1
/
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+
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Q.
In the expansion of
(
7
1
/
3
+
11
1
/
9
)
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prove that three will be only
730
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