# Binomial Theorem For Positive Integral Indices

INTRODUCTION:

In algebra, the binomial theorem shows the algebraic expansion for binomial with powers. by using  binomial theorem it is possible to expand the power (x + y) n into a sum containing terms of the form axbyc, where the coefficient of every  term is a positive, and the sum of the exponents of x and y in each term is n.The expansion of positive integral indices is using binomial theorem and the explanation for binomial theorem is given in the following sections

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## Binomial theorem:

With the help of this binomial theorem for positive integral index indices , we can expand any power of x + y into a sum of terms forming a polynomial. The terms are called as binomial co -efficients. Since it is only for positive integral indices,it expands only positive indices and not negative indices.

(x+y) n =       nc0. Xn.y0 + nc1 .x (n-1) y1 + nc2 .x (n-2).y2+……..+nc (n-1) x1.y (n-1) + ncn.x0.yn

Where the corresponding binomial theorem coefficient example is in the form ‘nCk.

nCk = `(n!) / [k! (n-k)!]`

## Example for expansion for positive integral indices:

Here are few examples for binomial theorem for positive integral indices,

Problem 1:

1. Expand (1+4x)6 using binomial theorem.

Binomial theorem [ (x+y) n =       nc0. Xn.y0 + nc1 .x (n-1) y1 + nc2 .x (n-2).y2+……..+nc (n-1) x1.y (n-1) + ncn.x0.yn     ]

Using theorem the positive integral indice can be expanded as,
(1+4x)6 =   6C0 . 16. (4x)0+ 6C1 15 (4x)1+ 6C2 1 (4x)2+ 6C3 13 (4x)3+ 6C4 12 (4x)4+ 6C11 (4x)5  +   6C10 (4x)6

= 1+ 5 (4x) +16x2 (10) + 64x(10) + 5 (256x) + 1024 x

= 1 + 20x + 160x2  + 640x3 + (640x) + 1024 x5

Problem 2:

Expand (x+y)3 using binomail theorem,

Binomial theorem [ (x+y) n =       nc0. Xn.y0 + nc1 .x (n-1) y1 + nc2 .x (n-2).y2+……..+nc (n-1) x1.y (n-1) + ncn.x0.yn     ]

(x+y)33C0 . x3. (y)0+ 3C1 x2 (y)1+ 3C2 x (y)2+ 3C3 x0 (y)3

=  1 (x3) + 3 ( x2 (y)1 ) + 3 ( x (y)2 ) + 1 ((y)3)

=   x3 + 3  x2 y  + 3  x  y2 ) +  y3

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## Practice problems:

Expand the following using binomial theorem for positive integral indices:

1. Expand (`sqrt(2)` + 1)6 + (`sqrt(2)` - 1)6                       ============> [Answer: 198]

2. Expand (10.1)5   [Hint: use 10.1 = 10 +0.1]   ============> [Answer: 105101.00501]