# What is an easy way to solve problems using the Binomial Theorem? Or could you explain how to do it?

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### 1 Answer

You may use binomial theorem, by itself or combined with other theorems, to solve different problems. For example, you may write the number "e", not only as a limit, but as a series, using binomial expansion, such that:

`(1 + 1/n)^n = C_n^0*1^n + C_n^1*1^(n-1)*(1/n) + C_n^2*1^(n-2)*(1/n)^2 + ... + C_n^k*1^(n-k)*(1/n)^k + .. + C_n^n*(1/n)^n`

The general "k+1"th term of this expansion is:

`T_(k+1) = C_n^k*1^(n-k)*(1/n)^k = (n!)/(k!(n-k)!)*1/(n^k)`

`T_(k+1) = 1/(k!)*1/(n^k)*(n!)/((n-k)!)`

`T_(k+1) = 1/(k!)*(n(n-1)...(n-k+1))/(n^k)`

Taking limit both sides, as n approaches to `oo` , yields:

`lim_(n->oo) T_(k+1) = lim_(n->oo) 1/(k!)*(n(n-1)...(n-k+1))/(n^k)`

`lim_(n->oo) T_(k+1) = 1/(k!) lim_(n->oo) (n(n-1)...(n-k+1))/(n^k)`

`lim_(n->oo) T_(k+1) = 1/(k!)*1`

Hence, evaluating the limitÂ `lim_(n->oo)(1 + 1/n)^n` means to evaluate each term of expansion, such that:

`lim_(n->oo)(1 + 1/n)^n = lim_(n->oo)C_n^0*1^n + lim_(n->oo)C_n^1*1^(n-1)*(1/n) + lim_(n->oo)C_n^2*1^(n-2)*(1/n)^2 + ... + lim_(n->oo)C_n^k*1^(n-k)*(1/n)^k + .. + lim_(n->oo)C_n^n*(1/n)^n`

`lim_(n->oo)(1 + 1/n)^n = 1/(0!) + 1/(1!) + 1/(2!) + ... + 1/(k!) + ... + ..`

`e = 1/(0!) + 1/(1!) + 1/(2!) + ... + 1/(k!) + ... + ..`

Hence, the number e may be written as it follows:

`e = 1/(0!) + 1/(1!) + 1/(2!) + ... + 1/(k!) + ... + .` .

The binomial theorem may be used, along with Moivre's theorem, to determine the formulas for double, triple or multiple angles.

Using Moivre's theorem for a complex number, raised to nth power, yields:

`(cos x + i*sin x)^n = cos (n*x) + i*sin(n*x)`

Using binomial theorem, yields:

`(cos x + i*sin x)^n = C_n^0*1(cos x)^n + C_n^1*(cos x)^(n-1)*(i*sin x) + C_n^2*(cos x)^(n-2)*(i*sin x)^2 + ... + C_n^k*(cos x)^(n-k)*(i*sin x)^k + .. + C_n^n*(i*sin x)^n`

Put n = 2, such that:

`(cos x + i*sin x)^2 = C_n^0*1(cos x)^2 + C_n^1*(cos x)^1*(i*sin x) + C_n^2*(cos x)^(0)*(i*sin x)^2`

`(cos x + i*sin x)^2 = cos^2 x + 2cos x*i*sin x + i^2*sin^2 x`

Since `i^2 = -1` , yields:

`(cos x + i*sin x)^2 = cos^2 x + 2cos x*i*sin x - sin^2 x`

Using Moivre's formula yields:

`(cos x + i*sin x)^2 = cos 2x + i*sin (2x)`

You need to set equal the real parts:

`cos^2 x - sin^2 x = cos 2x`

You need to set equal the imaginary parts:

`2cos x*sin x = sin 2x`

Hence, putting the real parts equal and imaginary parts equal yields the formulas for double angles.

**Hence, using the binomial theorem, you may solve numerous applications, in trigonometry, algebra and calculus.**