div10 wrote:

If \((a+b)^5=a^5+xa^4b+ya^3b^2+za^2b^3+tab^4+b^5\). What is the value of x+y+z+t?

A. 7

B. 15

C. 20

D. 30

E. 35

We can solve the problem by using the pascals triangle - The coefficients(the numbers in front of each term) follow a pattern

1

\((a+b)^0 = 1\)

1 1

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

1 2 1

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

1 3 3 1

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

1 4 6 4 1

\((a + b)4 = a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b4\)

1 5 10 10 5 1

\((a+b)^5 = a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+b^5\)

Therefore, Value of x+y+z+t = \(5+10+10+5 = 30\)(Option D)

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