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The Binomial Theorem — Easy Worksheet 2
Grade 12 · MathSheets Name: ____________________ Date: __________
Answer each question. Give coefficients as whole numbers; terms like 80x^2.
How to solve
(a + b)n = Σ C(n, k)·an-k·bk, where C(n, k) = n! / (k!(n - k)!) is the kth entry of row n of Pascal's triangle. To get a coefficient, pick the k that gives the power you want and multiply C(n, k) by the powers of the numbers (keep their signs).
Example
Find the coefficient of x3 y in the expansion of (x + y)4.
Answer: 4
Not quite. The answer is
💡 Each term of (x + y)4 is C(4, k) x4−k yk. For x2 y2 you need k = 2, so the coefficient is C(4, 2) = 4!/(2! · 2!) = 6. The answer is 6.
✓ Correct!
Not quite. The answer is
💡 Each term of (x + y)4 is C(4, k) x4−k yk. For x2 y2 you need k = 2, so the coefficient is C(4, 2) = 4!/(2! · 2!) = 6. The answer is 6.
💡 How to solve: The Binomial Theorem
(a + b)n = Σ C(n, k)·an-k·bk, where C(n, k) = n! / (k!(n - k)!) is the kth entry of row n of Pascal's triangle. To get a coefficient, pick the k that gives the power you want and multiply C(n, k) by the powers of the numbers (keep their signs).
Try a similar one
Find the coefficient of x² in the expansion of (x + 1)3.
✓ You've got it! Press Next to keep going. Not yet — the answer is . Read the steps above once more.
Not quite. The answer is
💡 Each term of (x + 2)5 is C(5, j) x5−j · 2j. For x you need j = 4. C(5, 4) = 5!/(4! · 1!) = 5, so the coefficient is C(5, 4) · 24 = 5 × 16 = 80.
✓ Correct!
Not quite. The answer is
💡 Each term of (x + 2)5 is C(5, j) x5−j · 2j. For x you need j = 4. C(5, 4) = 5!/(4! · 1!) = 5, so the coefficient is C(5, 4) · 24 = 5 × 16 = 80.
💡 How to solve: The Binomial Theorem
(a + b)n = Σ C(n, k)·an-k·bk, where C(n, k) = n! / (k!(n - k)!) is the kth entry of row n of Pascal's triangle. To get a coefficient, pick the k that gives the power you want and multiply C(n, k) by the powers of the numbers (keep their signs).
Try a similar one
Find the coefficient of x3 y2 in the expansion of (x + y)5.
✓ You've got it! Press Next to keep going. Not yet — the answer is . Read the steps above once more.
Not quite. The answer is
💡 Each term of (x + y)4 is C(4, k) x4−k yk. For x3 y you need k = 1, so the coefficient is C(4, 1) = 4!/(1! · 3!) = 4. The answer is 4.
✓ Correct!
Not quite. The answer is
💡 Each term of (x + y)4 is C(4, k) x4−k yk. For x3 y you need k = 1, so the coefficient is C(4, 1) = 4!/(1! · 3!) = 4. The answer is 4.
💡 How to solve: The Binomial Theorem
(a + b)n = Σ C(n, k)·an-k·bk, where C(n, k) = n! / (k!(n - k)!) is the kth entry of row n of Pascal's triangle. To get a coefficient, pick the k that gives the power you want and multiply C(n, k) by the powers of the numbers (keep their signs).
Try a similar one
Find the coefficient of xy2 in the expansion of (x + y)3.
✓ You've got it! Press Next to keep going. Not yet — the answer is . Read the steps above once more.
Not quite. The answer is
💡 Each term of (x + 3)3 is C(3, j) x3−j · 3j. For x you need j = 2. C(3, 2) = 3!/(2! · 1!) = 3, so the coefficient is C(3, 2) · 32 = 3 × 9 = 27.
✓ Correct!
Not quite. The answer is
💡 Each term of (x + 3)3 is C(3, j) x3−j · 3j. For x you need j = 2. C(3, 2) = 3!/(2! · 1!) = 3, so the coefficient is C(3, 2) · 32 = 3 × 9 = 27.
💡 How to solve: The Binomial Theorem
(a + b)n = Σ C(n, k)·an-k·bk, where C(n, k) = n! / (k!(n - k)!) is the kth entry of row n of Pascal's triangle. To get a coefficient, pick the k that gives the power you want and multiply C(n, k) by the powers of the numbers (keep their signs).
Try a similar one
Find the coefficient of x3 y in the expansion of (x + y)4.
✓ You've got it! Press Next to keep going. Not yet — the answer is . Read the steps above once more.
Not quite. The answer is
💡 Each term of (x + 1)6 is C(6, j) x6−j · 1j. For x4 you need j = 2. C(6, 2) = 6!/(2! · 4!) = 15, so the coefficient is C(6, 2) · 12 = 15 × 1 = 15.
✓ Correct!
Not quite. The answer is
💡 Each term of (x + 1)6 is C(6, j) x6−j · 1j. For x4 you need j = 2. C(6, 2) = 6!/(2! · 4!) = 15, so the coefficient is C(6, 2) · 12 = 15 × 1 = 15.
💡 How to solve: The Binomial Theorem
(a + b)n = Σ C(n, k)·an-k·bk, where C(n, k) = n! / (k!(n - k)!) is the kth entry of row n of Pascal's triangle. To get a coefficient, pick the k that gives the power you want and multiply C(n, k) by the powers of the numbers (keep their signs).
Try a similar one
Find the coefficient of xy5 in the expansion of (x + y)6.
✓ You've got it! Press Next to keep going. Not yet — the answer is . Read the steps above once more.
Not quite. The answer is
💡 Each term of (x + 2)4 is C(4, j) x4−j · 2j. For x³ you need j = 1. C(4, 1) = 4!/(1! · 3!) = 4, so the coefficient is C(4, 1) · 21 = 4 × 2 = 8.
✓ Correct!
Not quite. The answer is
💡 Each term of (x + 2)4 is C(4, j) x4−j · 2j. For x³ you need j = 1. C(4, 1) = 4!/(1! · 3!) = 4, so the coefficient is C(4, 1) · 21 = 4 × 2 = 8.
💡 How to solve: The Binomial Theorem
(a + b)n = Σ C(n, k)·an-k·bk, where C(n, k) = n! / (k!(n - k)!) is the kth entry of row n of Pascal's triangle. To get a coefficient, pick the k that gives the power you want and multiply C(n, k) by the powers of the numbers (keep their signs).
Try a similar one
Find the coefficient of xy4 in the expansion of (x + y)5.
✓ You've got it! Press Next to keep going. Not yet — the answer is . Read the steps above once more.
Not quite. The answer is
💡 Each term of (x + 2)3 is C(3, j) x3−j · 2j. For x you need j = 2. C(3, 2) = 3!/(2! · 1!) = 3, so the coefficient is C(3, 2) · 22 = 3 × 4 = 12.
✓ Correct!
Not quite. The answer is
💡 Each term of (x + 2)3 is C(3, j) x3−j · 2j. For x you need j = 2. C(3, 2) = 3!/(2! · 1!) = 3, so the coefficient is C(3, 2) · 22 = 3 × 4 = 12.
💡 How to solve: The Binomial Theorem
(a + b)n = Σ C(n, k)·an-k·bk, where C(n, k) = n! / (k!(n - k)!) is the kth entry of row n of Pascal's triangle. To get a coefficient, pick the k that gives the power you want and multiply C(n, k) by the powers of the numbers (keep their signs).
Try a similar one
Find the coefficient of x3 y in the expansion of (x + y)4.
✓ You've got it! Press Next to keep going. Not yet — the answer is . Read the steps above once more.
Not quite. The answer is
💡 Each term of (x + y)5 is C(5, k) x5−k yk. For xy4 you need k = 4, so the coefficient is C(5, 4) = 5!/(4! · 1!) = 5. The answer is 5.
✓ Correct!
Not quite. The answer is
💡 Each term of (x + y)5 is C(5, k) x5−k yk. For xy4 you need k = 4, so the coefficient is C(5, 4) = 5!/(4! · 1!) = 5. The answer is 5.
💡 How to solve: The Binomial Theorem
(a + b)n = Σ C(n, k)·an-k·bk, where C(n, k) = n! / (k!(n - k)!) is the kth entry of row n of Pascal's triangle. To get a coefficient, pick the k that gives the power you want and multiply C(n, k) by the powers of the numbers (keep their signs).
Try a similar one
Find the coefficient of x2 y2 in the expansion of (x + y)4.
✓ You've got it! Press Next to keep going. Not yet — the answer is . Read the steps above once more.
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The Binomial Theorem — Easy Worksheet 2
Answer each question. Give coefficients as whole numbers; terms like 80x^2.