Binomial Theorem Math Example 2

Follow the full solution, then compare it with the other examples linked below.

Example 2

hard
Find the coefficient of x3x^3 in the expansion of (2x+3)5(2x + 3)^5.

Solution

  1. 1
    Step 1: The general term is (5k)(2x)5−k(3)k\binom{5}{k}(2x)^{5-k}(3)^k.
  2. 2
    Step 2: For x3x^3, we need 5−k=35 - k = 3, so k=2k = 2.
  3. 3
    Step 3: Term = (52)(2x)3(3)2=10⋅8x3⋅9=720x3\binom{5}{2}(2x)^3(3)^2 = 10 \cdot 8x^3 \cdot 9 = 720x^3.
  4. 4
    Check: (52)=10\binom{5}{2} = 10, 23=82^3 = 8, 32=93^2 = 9, and 10×8×9=72010 \times 8 \times 9 = 720 ✓

Answer

720720
To find a specific term in a binomial expansion, identify the value of kk that gives the desired power of xx, then compute the binomial coefficient and remaining factors.

About Binomial Theorem

The binomial theorem gives the expansion of (a + b)^n as a sum of terms involving binomial coefficients: (a+b)^n = sum of C(n,k) * a^(n-k) * b^k. Each coefficient (nk)\binom{n}{k} counts the number of ways to choose kk copies of bb from nn factors.

Learn more about Binomial Theorem →

More Binomial Theorem Examples

Example 1 medium

Expand [formula] using the Binomial Theorem.

Example 3 easy

Expand [formula] using Pascal's triangle.

Example 4 medium

What is [formula]?

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