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BINOMIAL THEOREM Flipbook PDF
PART-I
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BINOMIAL THEOREM MANU
December 12, 2020
MANU
BINOMIAL THEOREM
n = 0:
1
n = 1:
1
n = 2:
1
n = 3: n = 4:
1 1
1 2
3 4
1 3
6
1 4
MANU
1
BINOMIAL THEOREM
BINOMIAL THEOREM
MANU
BINOMIAL THEOREM
BINOMIAL THEOREM a+b =a+b (a + b)2 = a2 + 2ab + b 2 (a + b)3 = a3 + 3a2 b + 3ab 2 + b 3 (a + b)4 = a4 + 4a2 b + 6ab 2 + 4ab 3 + b 4 (a + b)5 = a5 + 5a4 b + 10a3 b 2 + 10a2 b 3 + 5ab 4 + b 5 (a + b)6 = a6 + 6a5 b + 15a4 b 2 + 20a3 b 3 + 15a2 b 4 + 6ab 5 + b 6 (a + b)7 = a7 + 7a6 b + 21a5 b 2 + 35a4 b 3 + 35a3 b 4 + 21a2 b 5 + 7ab 6 + b 7
MANU
BINOMIAL THEOREM
BINOMIAL THEOREM
SPECIAL CASE Put a=1,b=x ,then (1 + x)n = nC0 + nC1 x + nC2 x 2 + ........ + nCn x n put x=1,2n = nC0 + nC1 + nC2 + ........ + nCn put a=1,b=-x,(1 − x)n = nC0 − nC1 x + nC2 x 2 − ........ + (−1)n nCn x n In the above put x=1, 0 = nC0 − nC1 + nC2 − ........ + (−1)n nCn MANU
BINOMIAL THEOREM
QUESTION01 Expand (1 + 2x)5 ANSWER01 (1 + 2x)5 = 1 + 5C1 x + 5C2 x 2 + 5C3 x 3 + 5C4 x 4 + x 5 = 1 + 5x + 10x 2 + 10x 3 + 5x 4 + x 5 QUESTION02 Expand (1 − 2x)5 Answer02 (1 − 2x)5 = 1 − 5C1 x + 5C2 x 2 − 5C3 x 3 + 5C4 x 4 − x 5 1 − 5x + 10x 2 − 10x 3 + 5x 4 − x 5
MANU
BINOMIAL THEOREM
QUESTION03 Expand (2x − 3)6 ANSWER03 (2x − 3)6 = (2x)6 − 6C1 (2x)5 (3)+ 6C2 (2x)4 (3)2 − 6C3 (2x)3 (3)3 + 6C4 (2x)2 (3)4 − 6C5 2x(3)5 +36 = 64x 6 − 6.32.3x 5 + 15.16.9x 4 − 20.8.27x 3 + 15.4.81x 2 − 6.2.243x + 729 = 64x 6 − 576x 5 + 2160x 4 − 4320x 3 + 4860x 2 − 2916x + 729 QUESTION04 Expand ( x2 − x2 )5 ANSWER04 ( x2 − x2 )5 = ( x2 )5 − 5C1 ( x2 )4 ( x2 ) + 5C2 ( x2 )3 ( x2 )2 − 5C3 ( x2 )2 ( x2 )3 + 5C4 ( x2 )( x2 )4 − ( x2 )5 =
32 x5
− 5 x164
=
32 x5
−
40 x3
x 2
+
+ 10 x83 20 x
x2 4
− 5x +
− 10 x42 5x 3 8
−
x3 8
+ 5 x2
x4 16
−
x5 32
x5 32
MANU
BINOMIAL THEOREM
QUESTION05 Expand ( x3 + x1 )5 ANSWER05 ( x3 + x1 )5 = ( x3 )5 + 5C1 ( x3 )4 x1 + 5C2 ( x3 )3 ( x1 )2 + 5C3 ( x3 )2 ( x1 )3 + 5C4 ( x3 )( x1 )4 + =
5
x 243
+
3
5x 81
+
10x 27
+
10 9x
+
5 3x 3
+
1 x5
QUESTION06 Using Binomial theorem Expand(102)5 Answer06 1025 = (100 + 2)5 1005 + 5C1 1004 .2 + 5C2 1003 .22 + 5C3 1002 .23 + 5C4 100.24 + 25 = 10000000000 + 1000000000 + 40000000 + 800000 + 8000 + 32 = 11040808032
MANU
BINOMIAL THEOREM
1 x5
QUESTION07 Using Binomial theorem Expand(99)5 ANSWER07 995 = (100 − 1)5 1005 − 5C1 1004 .1 + 5C2 1003 .12 − 5C3 1002 .13 + 5C4 100.14 − 15 = 10000000000 − 500000000 + 10000000 − 100000 + 500 − 1 = 10010000500 − 500100001 = 9509900499 QUESTION08 Using Binomial Theorem, indicate which number is larger (1.1)10000 or 1000. ANSWER08 (1.1)10000 = (1 + .1)10000 1 + 10000C1 19999 .1 + 10000C2 19998 (.1)2 + .......... 1 + 1000 + postive numbers........ > 1000
MANU
BINOMIAL THEOREM
QUESTION09 √ √ √ √ Find (x + y )4 − (x − y )4 Hence evaluate ( 5 + 6)4 − ( 5 − 6)4 ANSWER09 (x + y )4 − (x − y )4 = (x 4 + 4C1 x 3 y + 4C2 x 2 y 2 + 4C3 xy 3 + y 4 ) − 4 4 (x − C1 x 3 y + 4C2 x 2 y 2 − 4C3 xy 3 + y 4 ) (x y )4 − (x − y )4 = + (x 4 + 4x 3 y + 6x 2 y 2 + 4xy 3 + y 4 ) − (x 4 − 4x 3 y + 6x 2 y 2 − 4xy 3 + y 4 ) = 8x 3 y + 8xy 3 = 8xy (x 2 + y 2 ) √ √ √ √ ( 5 + 6)4 − ( 5 − 6)4 √ √ h √ √ i = 8 5 6 ( 5)2 + ( 6)2 √ = 88 30
MANU
BINOMIAL THEOREM
QUESTION10 √ √ Find (x + 1)6 + (x − 1)6 Hence evaluate ( 2 + 1)6 + ( 2 − 1)6 ANSWER10 (x + 1)6 + (x − 1)6 = (x 6 + 6C1 x 5 + 6C2 x 4 + 6C3 x 3 + 6C4 x 2 + 6C5 x + 1) + 6 6 (x − C1 x 5 + 6C2 x 4 − 6C3 x 3 + 6C4 x 2 − 6C5 x + 1) = 2(x 6 + 6C2 x 4 + 6C4 x 2 + 1) = 2(x 6 + 15x 4 + 15x 2 + 1) √ √ ( 2 + 1)4 + ( 2 − 1)6 √ √ √ = 2(( 2)6 + 15( 2)4 + 15 2)2 + 1) = 2(8 + 60 + 30 + 1) = 2 × 99 = 198
MANU
BINOMIAL THEOREM
QUESTION11 Show that 9n+1 − 8n − 9 is divisible by 64,where n is a postive integer ANSWER11 9n+1 = (1 + 8)n+1 = 1 + n+1C1 8 + n+1C2 82 + ..... + 8n+1 = 1 + (n + 1)8 + 82 n+1C2 + .........8n−1 9n+1 − 1 − 8(n + 1) = 64 n+1C2 + .........8n−1 9n+1 − 8n − 9 = M(64) Hence 9n+1 − 8n − 9 is divisible by 64
MANU
BINOMIAL THEOREM