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Published on: 16/09/2019
Binomial Theorem
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1.
In the binomial expansion of (1 + x)n, the coefficient of the 5th, 6th and 7th terms are in A.P. Find all values of n for which this can happen.
2.
Find the value of a for which the co-efficient of the middle term in the expansions of (1 + ax)4 and (1 - ax)6 are equal.
3.
Find the co-efficient of x6y3 in (x + 2y)9
4.
Find the co-efficient of \({ x }^{ 10 }\quad in\quad \left( 2x^{ 2 }-\frac { 3 }{ x } \right) ^{ 11 }\)
5.
Find (a + b)5 - (a - b)5. Hence evaluate \(\left( \sqrt { 5 } +\sqrt { 3 } \right) ^{ 5 }+\left( \sqrt { 5 } -\sqrt { 3 } \right) ^{ 5 }\)
6.
Expand the following expressions by using binomial theorem:
(1- 3x)7
7.
If the coefficients of x7 and x8in the expansion of \(\left( 3+\frac { x }{ { 2 } } \right) ^{ n }\)
8.
Find the term independent of x in the expansion of \(\left( 2x+\frac { 1 }{ { 3x }^{ 2 } } \right) ^{ 9 }\)
9.
Find the middle term in the expansion of \(\left( \frac { x }{ a } -\frac { a }{ x } \right) ^{ 10 }\)
10.
Find the value of [x+\(\sqrt { x-1 } \)]6+[\(\sqrt { x-1 } \)]6 , using binomial theorem.
11.
Find the number of terms in the expansion of (1 + 2x + x2)14
12.
using bionominal theorem compute (10.1)5.
13.
Using binomial theorem, expand \( \left( { x+ }\frac { 1 }{ y } \right) ^{ 6 }\)
14.
If P and Q are the sum of odd and even terms in the expansion\(\left( y+b \right) ^{ n }\), then prove that\(\left( y+b \right) ^{ 2n }+\left( y-b \right) ^{ 2n }=2\left( { P }^{ 2 }+Q^{ 2 } \right) \).
15.
Find the coefficient of x3 in the expansion of \(\left( 3x-\frac { 1 }{ x } \right) ^{ 7 }\)
1.
7, 14
2.
\(a=\frac{-3}{10}\)
3.
672
4.
3421440
5.
2(5a4b + 10a2b3 + b5); 568\(\sqrt3\)
6.
1- 21x + 189x2 - 945x3+ 2835x4 - 5103x5+ 5103x6 -2187x7
7.
55
8.
\(\frac { 1792 }{ 9 } \)
9.
-252
10.
2 (x6+ 15x5 - 29x3 + 12x2 + 3x - 1)
11.
Here(1 + 2x + x2)14
=[(1 + x)2]14 = (1 + x)28
The number of terms in the expansion of
(1 + 2x + X2)14= 28 + 1 = 29
12.
using bionominal theorem for the expansion of (10.1)5, we have
(10.5)5 = (10 + 0.1)5 = 5C0(!0)5 + 5C1(10)4(0.1) + 5C2(10)3(0.1)2 + 5C3(10)2(0.1)5 + 5C4(10)(0.1)4 + 5C5(0.1)5
= (10)5 + 5(10)4(0.1) + 10(10)3(0.1)2 + 10(10)2(0.1)3 + 5(10)(0.1)4 + (0.1)5
= 100000 + 5000 + 100 + 1 + 0.005 + 0.00001 = 105101.00501
13.
Using binomial theorem for the expansion of \(\left( { x+ }\frac { 1 }{ y } \right) ^{ 6 }\) , we have
\(\left( { x+ }\frac { 1 }{ y } \right) ^{ 6 }\) = \(^{ 6 }{ c }_{ 0 }x^{ 6 }+^{ 6 }{ c }_{ 1 }{ x }^{ 5 }\left( \frac { 1 }{ y } \right) +^{ 6 }{ c }_{ 2 }{ x }^{ 4 }\left( \frac { 1 }{ y } \right) ^{ 2 }+^{ 6 }{ c }_{ 3 }{ x }^{ 3 }\left( \frac { 1 }{ y } \right) ^{ 3 }+^{ 6 }{ c }_{ 4 }{ x }^{ 2 }\left( \frac { 1 }{ y } \right) ^{ 4 }+^{ 6 }{ c }_{ 5 }{ x\left( \frac { 1 }{ y } \right) ^{ 5 }+^{ 6 }{ c }_{ 6 }\left( \frac { 1 }{ y } \right) }^{ 6 }\)
= \( { x }^{ 6 }+6\frac { { x }^{ 5 } }{ y } +15\frac { { x }^{ 4 } }{ { y }^{ 2 } } +20\frac { { x }^{ 3 } }{ { y }^{ 3 } } +15\frac { { x }^{ 2 } }{ { y }^{ 4 } } +6\frac { { x } }{ { y }^{ 5 } } +\frac { 1 }{ { y }^{ 6 } } \)
14.
\(\left( y+b \right) ^{ n }=^{ n }{ { C }_{ 0 }{ y }^{ n } }+^{ n }{ { C }_{ 1 }{ y }^{ n-1 }b+^{ n }{ { C }_{ 2 }{ y }^{ n-2 }{ b }^{ 2 }+.... } }+^{ n }{ { C }_{ n }{ b }^{ n } }\)
\(=P+Q\) .....(i)
\( \left( y-b \right) ^{ n }=^{ n }{ { C }_{ 0 }{ y }^{ n } }-^{ n }{ { C }_{ 1 }{ y }^{ n-1 }b+^{ n }{ { C }_{ 2 }{ y }^{ n-2 }{ b }^{ 2 }-...... } }+^{ n }{ { C }_{ n }{ b }^{ n } }\)
\(=P+Q\) ...(ii)
Squaring Eqs. (i) and (ii) then adding, we get
\(\left( y+b \right) ^{ 2n }+\left( y-b \right) ^{ 2n }={ \left( P+Q \right) }^{ 2 }+{ \left( P-Q \right) }^{ 2 }\)
\(={ P }^{ 2 }+Q^{ 2 }+2PQ+{ P }^{ 2 }+Q^{ 2 }-2PQ=2\left( { P }^{ 2 }+Q^{ 2 } \right) \)
15.
\(T_{ r+1 }={ ^{ 7 }C }_{ r }\quad 3^{ 7-r }\times \quad x^{ 7-2r }\quad (-1)^{ r }\)
For coefficient of x3, put 7-2r=3 \(\Rightarrow \) r=2
\({ T }_{ 3 }=T_{ 2+1 }={ ^{ 7 }C }_{ r }\quad 3^{ 7-2 }\quad x^{ 7-4 } =5103\)
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