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The binomial theorem is an algebraic method through which we can discover the price of any integer exponent of a binomial of the shape x y as a polynomial of the nth diploma of x and y. When a binomial is inside the extra shape, the simplest multiples of the time period are the binomial factors. We can discover the rectangular and dice of phrases like a b and a – b. The binomial theorem offers us an easy manner to simplify huge phrases.
The Maclaurin series, commonly known as the binomial series, is unique in mathematics.
This series is a variant of the binomial theorem, with a=1 and b=x. The series can be converging or diverging depending on the values of a and n.
(1 + a)n = nC0 + nC1a + nC2a2 + …… nCrar + …. + nCnan
Important Question
Determine the unbiased time period of y in the growth of (3y – (2/y2))15
Solution:
Given us,
The general term of (3y – (2/y2))15
and T(a + 1) = 15Ca (3y)15-a (-2/y2)a. It is independent of y if,
15 – a – 2a = 0 => a = 5
.·. T6= 15C5(3)10(-2)5 = – 16C5 310 25.
Determine the value of “a” if the coefficients of the (2a + 4)th and (a – 2)th terms in the (1 + y)18 expansion are equivalent.
Solution:
The general term of (1 + y)n is Ta + 1 = Cayr
Hence coefficient of (2a + 4)th term will be
T2a + 4 = T2a + 3 + 1 = 18C2a + 3
and coefficient or (a – 2)th term will be
Ta – 2 = Ta – 3 + 1 = 18Ca – 3.
=> 18C2a + 3 = 18Ca – 3.
=> (2a + 3) + (a – 3) = 18 (·.· nCa = nCK => a = k or a + k = n)
.·. a = 6
In the expansion of 3(1 + (1/3)) 20, find the greatest term’s value.
Solution:
Let Ta + 1 be the greatest term, then Ta < Ta + 1 > Ta + 2
Consider : Ta + 1 > Ta
=> 20Ca (1/√3)a > 20Ca – 1(1/√3)a-1
=> ((20)!/(20 – a)!a!) (1/(√3)a) > ((20)!/(21 – a)!(a – 1)!) (1/(√3)a-1)
=> a < 21/(√3 + 1)
=> a < 7.686 ……… (i)
Similarly, considering Ta + 1 > Ta + 2
=> a > 6.69 ………. (ii)
From (i) and (ii), we get
a = 7
Hence greatest term T8 = 25840/9
If (15 + 6√6)2n+1 = A, then A (1 – X) = 92n + 1 (where X is the fractional part of A).
Solution:
We can write
A = (15 + 6√6)2n+1 = I + X (Where I is integral and X is the fractional part of A)
Let X’ = (15 + 6√6)2n+1
6√6 = 14.69
=> 0 < 156√6 < 1
=> 0 < (15 + 6√6)2n+1 < 1
=> 0 < F’ < 1
Now, I + X = C0 (15)2n+1 + C1(15)2n 6√6 + C2 (15)2n (6√6 )2 +…
X’ = C0 (15)2n+1 – C1(15)2n 6√6 + C2(15)2n-1 (6√6 )2 +…
I + X + X’ = 2[C0 (15)2n+1 + C2 (15)2n-1 (6√6 )2 +…]
Term on R.H.S. is an even integer.
=> I + X + X’ = Even integer
=> X + X’ = Integer
But, 0 < X < 1 and (X is fraction part)
0 < X’ < 1
=> 0 < X + X’ < 2
Hence X + X’ = 1
X’ = (1 – x)
.·. A(1 – X) = (15 + 6√6 )2n+1 (15 – 6√6 )2n+1
= (9)2n+1 (Hence, proved)
Prove that if any four consecutive terms in the expansion of (1 + a)n have coefficients x1, x2, x3, and x4, then
x1/(x1 + x2) + x2/(x3 + a4) = 2 x 2/(x2 + x3)
Solution:
Because the coefficients x1, x2, x3, and x4 are coefficients of consecutive phrases,
Let x1 = nCr
x2 = nCr + 1
x3 = nCr + 2 and
x4 = nCr + 3
Now x1/(x1 + x2) = nCr/(nCr + nCr + 1) = 1/(1+ ((n – r)/(r + 1))) = (r + 1)/(n + 1)
Similarly, x2/(x2 + x3) = (r + 3)/(n + 1)
Now x3/(x3 + x4) + x1/(x1 + x2) = (2r + 4)/(n + 1)
= 2(r + 1)/(n + 1) = 2 x 2/(x2 + x3) (Hence, proved)
Determine which is the larger.9950 + 10050 or 10150.
Solution:
Let’s try to find out 10150 – 9950 in terms of remaining term i.e.
10150 – 9950 = (100 + 1)50 – (100 – 1)50
= (C0.10050 + C110049 + C2.10048 +……)
= (C010050 – C110049 + C210048 -……)
= 2[C1.10049 + C310047 +………]
= 2[50.10049 + C310047 +………]
= 10050 + 2[C310047 +…………]
> 10050
=> 10150 > 9950 + 10050
If x and m are both positive integers, thenand
Sx = 1x + 2x +………+ nx
Then demonstrate that
n+1C1S1 + n+1C2S2 +…+ n+1CnSn = (1 + m)n+1 – (1 + m)
Solution:
We have,
(1 + a)n+1 = n+1C0 + n+1C1 a + n+1C2 a2 + n+1Cn + 1 an+1
Putting a = 1, 2, 3, ………, m
=>2n+1 – 1 = n+1C0 + n+1C1(1) + n+1C2(1)2 +…+ n+1Cm(1)m
3n+1 – 2n+1 = n+1C0 + n+1C1(2) + n+1C2(2)2 +…+ n+1Cn(2)n
4n+1 – 3n+1 = n+1C0 + n+1C1(3) + n+1C2(3)2 +…+ n+1Cn(3)n
(1 + m)n+1 – mn+1 = n+1C0(m) + n+1C2 Σm + n+1C2 Σ(m)2 +…+ n+1Cn ∑(m)n.
Adding all these terms, we get
(1 + m)n+1 – 1 = n+1C0(m) + n+1C1S1 + n+1C2S2 +…+ n+1CnSn
=>n+1C1S1 + n+1C2S2 +…+ n+1CnSn = (1 + m)n+1 – (1 + m)
(Hence, proved)
In the expansion, find the (y)50coefficient of (1 + y)1000 + 2y(1 + y)999 + 3y2 (1 + y)998 +…+ 1001y1000.
Solution:
Let A = (1 + y)1000 + 2y(1+y)999 +…+ 1000y999 (1+y) + 1001 y1000
This is an Arithmetic Geometric Series with r = y/(1+y) and d = 1.
Now (y/(1+y)) S = y(1 + y)999 + 2y2 (1 + y)998 +…+ 1000y1000 + 1000y1001/(1+y)
Subtracting we get,
(1 – (y/(y + 1))) A = (1 + y)1000 + y(1 + y)999 +…+ y1000 – 1001 y1000/(1 + y)
or A = (1 + y)1001 + y (1 + y)1000 + y2 (1 + y)999 +…+ y1000 (1 + y) – 1001y1001
This is G.P. and the sum is
A = (1 + y)1002 – y1002 – 1002y1001
Then, coefficient of y50 is = 1002C50
Conclusion
The binomial theorem is commonly used in statistical and stochastic analysis. It is very convenient because our economy relies on statistical and probabilistic analysis. High-level mathematics and calculations use the binomial theorem to find the roots of higher power equations. Economists used the binomial theorem to calculate probabilities based on several widely used factors and predicted how the economy would develop in the coming years.
