Binom of Newton

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Let's write expressions for (a+b)^n when n=1,2,3,4:

(a+b)^1=\ \ \ \ \ \ \ \ \ \ \ \ \ a+b;

(a+b)^2=\ \ \ \ \ \ \ \ a^2+2ab+b^2;

(a+b)^3=\ \ \ \ a^3+3a^2b+3ab^2+b^3;

(a+b)^4=a^4+4a^3b+6a^2b^2+4ab^3+b^4.

We can notice some pattern here: when we raise binom to n-th power, we obtain n+1 summands; each summand contains multipliers a and b, sum of whose exponents adds up to n. Exponents of a decrease from n to 0, and exponents of b increase from 0 to n. Coefficients near powers of a and b in the expansion of (a+b)^n are called binomial coefficients. We can easily notice that binomial coefficients in the expansion of (a+b)^n are number of combinations from n elements of k=0,1,2,3,...n elements.

For example, in expansion of (a+b)^4 we have C_4^0=1,C_4^1=4,C_4^2=6,C_4^3=4,C_4^4=1.

At the beginning of this note I wrote expansion for n=1,2,3,4. Do you see how right-hand sides form Pascal's triangle?

Newton's Binom Formula. (a+b)^n=a^n+C_n^1a^(n-1)b+C_n^2a^(n-2)b^2+...+C_n^ka^(n-k)b^k+...+C_n^(n-1)ab^(n-1)+b^n.

Example. Write expansion of (1/x+2x^2)^5.

Here a=1/x,b=2x^2,n=5, therefore (1/x+2x^2)^5=(1/x)^5+C_5^1(1/x)^4*2x^2+C_5^2(1/x)^3(2x^2)^2+C_5^3(1/x)^2(2x^2)^3+

+C_5^4 1/x (2x^2)^4+(2x^2)^5=(1/x)^5+5(1/x)^4*2x^2+10(1/x)^3(2x^2)^2+10(1/x)^2(2x^2)^3+

+5 1/x (2x^2)^4+(2x^2)^5=1/x^5+10/x^2+40x+80x^4+80x^7+32x^10.