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Binomial Theorem Part 3
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Section:
Mathematics
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© examsnet.com
Question : 8
Total: 36
Let
(
1
+
x
+
x
2
)
10
=
a
0
+
a
1
x
+
a
2
x
2
+
.
.
.
+
a
20
x
20
. If
(
a
1
+
a
3
+
a
5
+
.
.
.
+
a
19
)
−
11
a
2
=
121
k
, then
k
is equal to
_
_
_
_
.
[3 Apr 2025 Shift 2]
Your Answer:
Validate
Solution:
Let
f
(
x
)
=
(
1
+
x
+
x
2
)
10
=
20
∑
r
=
0
a
r
x
r
The sum of odd coefficients:
S
‌
odd
‌
=
a
1
+
a
3
+
a
5
+
.
.
.
+
a
19
Subtracting
11
a
2
from above will give the answer
‌
S
‌
odd
‌
=
‌
f
(
1
)
−
f
(
−
1
)
2
‌
f
(
1
)
=
(
1
+
1
+
1
)
10
=
3
10
‌
f
(
−
1
)
=
(
1
−
1
+
1
)
10
=
(
1
)
10
=
1
‌
S
‌
odd
‌
=
∑
‌
odd
‌
r
a
r
=
‌
3
10
−
1
2
Now for
a
2
Now use:
‌
(
1
−
x
3
)
10
=
10
∑
x
=
0
(
−
1
)
k
(
10
k
)
x
3
k
‌
(
1
−
x
)
−
10
=
∞
∑
r
=
0
(
r
+
9
9
)
x
r
So
f
(
x
)
=
(
10
∑
k
=
0
(
−
1
)
k
(
10
k
)
x
3
k
)
â‹…
(
∞
∑
r
=
0
(
r
+
9
9
)
x
r
)
Only the term with
x
0
from the first sum (i.e.,
k
=
0
) can contribute to
x
2
, since all other
k
≥
1
gives
x
3
k
≥
x
3
From
(
1
−
x
3
)
10
: the
x
0
term is
(
10
0
)
=
1
From
(
1
−
x
)
−
10
: the coefficient of
x
2
is
(
2
+
9
9
)
=
(
11
9
)
=
55
Hence,
a
2
=
1.55
=
55
Now,
S
‌
odd
‌
−
11
a
2
=
‌
3
10
−
1
2
−
11
â‹…
55
=
121
k
3
10
=
59049
So:
‌
S
=
‌
59049
−
1
2
−
605
=
‌
59048
2
−
605
‌
=
29524
−
605
=
28919
So:
121
k
=
28919
⇒
k
=
‌
28919
121
=
239
© examsnet.com
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