1.Algebra Booster

Binomial Theorem 6.3
If r is a fraction, the next integer is numerically the greatest
term.
Alternative method
To determine the greatest term in the expansion of (a + x) n ,
we should use the following steps.
( n+
1)| x|
1. Calculate m = .
a + | x|
2. If m ΠI, then t m
and t m+1
are numerically the greatest
terms.
3. If m ΠI, then t |m|+1
is numerically the greatest term,
where [,] = GIF.
4.7 Divisibility-related Problems in Binomial
Theorem
Consider the expansion
(1 + b) n 2 3
= (1 + n Cb+ n C b + n Cb + +
n C b
n )
and so
1 2 3
n n n 2 n 3 n n
1 2 3
n
(1 + b) - 1 = ( C b+ C b + C b + + C b )
[(1 + b) n – 1] is divisible by b
Also [(1 + b) n – bn – 1] is divisible by b 2 and so on.
Divisibility of (a n ± b n ) by (a ± b), where a, b, n Œ N
(i) When n is even, (a n – b n ) is divisible by (a + b) and (a –
b)
(ii) When n is odd, (a n – b n ) is divisible by (a – b)
(iii) When n is odd, (a n + b n ) is divisible by (a + b).
4.8 Unit Digit of a Natural Number
The unit digit of a natural number depends on the period of
that number.
We have, 2 1 = 2, 2 2 = 4, 2 3 = 8, 2 4 = 16
Also, 2 5 = 32, 2 6 = 64, 2 7 = 128, 2 8 = 256
Thus, the period of 2 is 4
Similarly, the period of 3 is 4 and also the period of 7 is 4.
For example,
(i) the unit digit of 2 2013 = 2 (4¥503+1) is 2.
(ii) the unit digit of 2012 2013 is same as 2 2013 .
(iii) the unit digit of 3 2014 = 3 (4¥504+2) is 9.
(iv) the unit digit of (27) 50 = 2 150 = 3 (4¥37+2) is 9.
(v) The unit digit of 9 99 is 9, since the period of 9 is 2.
4.9 Rational Terms in the Expansion of (a 1/p , b 1/q ) n ,
where a and b are Prime Numbers, p and q
are Integers and n ΠN.
First of all, we find the (r + 1)th term in the expansion of (a 1/p
+ b 1/q ) n .
Therefore, t r+1
= n C r
(a 1/p ) n–r (b 1/q ) r
n
n-r r
p q
= Cr
¥ ( a) ( b)
Now, by inspection, putting the values of 0 £ r £ n, when
indices of a, b are integers.
n
4.10 Integral and Fractional part of a number
n
(i) If ( P + Q) = I + f,
where I and n are natural numbers,
n being odd and 0 £ f < 1,
(1 + f)f = k n , where P – Q 2 = k and Q - P< 1 .
n
(ii) If ( P + Q) = I + f,
where I and n are natural numbers,
n being even and 0 £ f < 1,
(I + f)(1 – f) = k n , where P – Q 2 = k and Q - P< 1 .
4.11 Properties of Binomial Co-efficients
Let n C 0
, n C 1
, n C 2
, …, n C n
are the binomial co-efficients in the
expansion of (1 + x) n .
P 1. n C0+ n C n n 1+ C2+ + C 2
n
n =
P 2.
n n n n-1
C + C + C + =
0 2 4 2
n n n n-
C1+ C3+ C5+ = 2
P 3.
1
P 4. n
C r
= n C n–r
n n
P 5. C = C fi x= yor
x+ y = n
P 6.
n
x
1
y
Cr
Ên- r + 1ˆ
= Á
C Ë
˜
-
r ¯
n
r
P 7. n
C r
+ n C n–r
= n+1 C r
n n 1 n 2 n 3 r n 1
P 8. C + - C + - C + - C + + C =
+ C
r r r r r r+
1
P 9. Summation upto middle term
2n 2n 2n 2n 2n
C0+ C1+ C2+ C3+ + Cn
2n-1 1 2n
= 2 + Cn
2
P 10. Summation of series
n
C 3
+ 2 n+1 C 3
+ 3 n+2 C 3
+ … + n 2n–1 C 3
= – 2n C 5
+ n C 5
+ n 2n C 4
P 11
n n n-1 n( n-
1) n-2
Cr = ◊ Cr-1= ◊ C and so on.
r-2
r r( r - 1)
P 12. Sum of the series, when the sum of the lower suffices
are the same
m C ◊ n C + m C ◊ n C + m C ◊
n C
r 0 r-1 1 r-2 2
m n m+
n
+ + Cm◊ Cr = Cr
P 13. Sum of the series, when the difference of the lower suffices
are the same
n n n n n n
C ◊ C + C ◊ C + C ◊ C
r n r-1 n-1 r-2 n-2
n n 2n
+ + C0
◊ Cn-r = Cn-r
5. MULTINOMIAL THEOREM FOR POSITIVE INTEGRAL INDEX
+
For nŒI
,
n
( a1+ a2+ a3+ + am
)
Ê n ˆ
1 2 k3
= Â Á a1 a2 a3
… a
Ëk , k , k ,…, k ˜
¯
k1+ k2+ k3+ + km
= n
1 2 3
m
k k km
m