What is the coefficient of the x^4y^3 term in the expansion of (x-2y) ^7

The coefficients are therefore

1,1

1,2,1

1,3,3,1

1,4,6,4,1

1,5,10,10,5,1

1,6,15,20,15,6,1

1,7,21,35,35,21,7,1

Step-by-step explanation:

(a+b) ^n

=a^n+na^ (n-1) b + (n (n-1) / 2) a^ (n-2) b² + ... + nab^ (n-1) + b^n

=C (n, 0) a^n+C (n, 1) a^ (n-1) b+C (n-2) a^ (n-2) b^sup2; ... C (n, n) b^n

For n=1

(a+b) 1=a+b

(a+b) 2=a²+2ab+b^sup2;

(a+b) 3=a³+3a^sup2; b+3ab²+b³

...

The coefficients are therefore

1,1

1,2,1

1,3,3,1

1,4,6,4,1

1,5,10,10,5,1

1,6,15,20,15,6,1

1,7,21,35,35,21,7,1

...

This is the Pascal's triangle

The sum of each pair of adjacent numbers gives rise to a number on the next line.

This also means that

(a+b) 7

=a^7+7a^6b+21a^5b²+35a^4b³ + ... + 7ab^6+b^7 ... (A)

which is almost the solution to the problem.

To solve the problem using the Pascal's triangle, substitute a=x, and b=2y in equation (A) to give the expression of the final answer.