1.Algebra Booster

Binomial Theorem 6.27
n
Now, ( P + Q) -( P - Q)
= 1 + f – f ¢
Since n is odd, the LHS of the above expression contains
an even powers of P.
Hence, the terms in LHS and I are integers.
Therefore, f – f¢is an integer.
But 0 < f < 1 and –1 < f¢ < 0
fi 0 < f < 1 and –1 < –f¢ < 0
fi –1 < f – f¢ < 1
fi f – f¢ = 0
fi f = f¢
Thus, ( I + f) f ¢ = (
n
P + Q) (
n
P -Q)
= {( P + Q)( n
P -Q)}
= (P – Q 2 ) n
= k n
51. Given Q - P<
1
fi 0 < ( Q - P) < 1
n
n
Let ( Q - P) = f ¢ , where 0 < f¢ < 1.
Now, ( P + Q) n + (
n
P - Q)
= I + f + f ¢
= Even integer
But 0 < f < 1 and 0 < f¢ < 1
fi 0 < f + f¢ < 2
fi f + f¢ = 1
fi f¢ = 1 – f
Hence,
(I + f)(I – f) = (I + f)f¢
= ( P + Q) n ¥ (
n
P -Q)
= (P – Q 2 ) n
= k n
52 Let I and f denote the integral and fractional part of R,
respectively.
Given f = R – [R]
2n+
1
Let R = I + f = (5 5 + 11) , 0 < f < 1
2n+
1
Also, f ¢ = (5 5 -11) , 0 < f¢ < 1
Now,
n
R – f¢ = (5 5 + 11) - (5 5 -11)
= an even integer.
But 0 < f < 1 and 0 < f¢ < 1
fi 0 < f < 1 and –1 < –f¢ < 0
fi –1 < f – f¢ < 1
fi f – f¢ = 0
fi f = f¢
Therefore, Rf = Rf¢
n
= (5 5 + 11) ◊(5 5 -11)
= (125 – 121) 2n+1
= 4 2n+1
n
2 + 1 2n+
1
2 + 1 2n+
1
53. Let x = I + f = (8 + 3 7) , where 0 < f < 1,
n
and f ¢ = (8 - 3 7) , 0 < f ¢ < 1.
n
Now,
1 + f + f¢ (8 3 7) n
n
= + + (8 -3 7)
= an even integer.
But 0 < f < 1 and 0 < f¢ < 1
fi 0 < f + f ¢ < 2
fi f + f ¢ = 1
Therefore, I + f + f ¢ = 2k
fi I + 1 = 2k
fi I = 2k – 1
fi [x] = 2k – 1
Now,
x – x 2 + x[x]=x – x(x – [x])
= x – xf
= x(1 – f)
= xf¢
n
n
= (8 + 3 7) ◊(8 -3 7)
= (64 – 63) n
= 1 n = 1
n
54. Given a + b = (8 + 3 7) , where 0 < b < 1.
n
Let b¢ = (8 - 3 7) , where 0 < b¢ < 1 .
Now,
a + b + b ¢ (8 3 7) n
n
= + + (8 -3 7)
= an even integer.
Also, 0 < b < 1 and 0 < b ¢ < 1
fi 0 < b + b ¢ < 2
fi b + b ¢ = 1
fi b ¢ = 1 – b
Thus,
(1 – b)(a + b) (8 3 7) n
n
= - ◊ (8 + 3 7)
= (64 – 63) n
= 1 n = 1
5
55. Let I + f = (3 + 5) , where 0 < f < 1,
5
and f ¢ = (3 - 5) , where 0 < f ¢ < 1 .
Now,
5 5
I + f + f¢ = (3 + 5) + (3 - 5)
5 5 5 3 5 1 2
= 2( C0 ◊ 3 + C2 ◊3 ◊ 5 + C4
◊3 ◊5 )
= 2(243 + 10.135 + 5.75)
= 2(243 + 1350 + 275)
= 3936
5
Thus, the number just grater than the number (3 + 5)
is I + f + f ¢ = 3936.
56. The co-efficient of a n in (1 + a) m+n
= m+n C n
= m+n C m+n–n
= m+n C m
= the co-efficient of a m in (1 + a) m+n .