A brief note on Properties and Applications of Binomial Coefficients . Read it before a test. Try each question at the end before you open its answer.
The rules What Rule symmetry n C r = n C n − r {}^{n}C_{r} = {}^{n}C_{n - r} n C r = n C n − r Pascal's rule n C r − 1 + n C r = n + 1 C r {}^{n}C_{r - 1} + {}^{n}C_{r} = {}^{n + 1}C_{r} n C r − 1 + n C r = n + 1 C r neighbours n C r n C r − 1 = n − r + 1 r \dfrac{{}^{n}C_{r}}{{}^{n}C_{r - 1}} = \dfrac{n - r + 1}{r} n C r − 1 n C r = r n − r + 1 a factor r r r r ⋅ n C r = n ⋅ n − 1 C r − 1 r \cdot {}^{n}C_{r} = n \cdot {}^{n - 1}C_{r - 1} r ⋅ n C r = n ⋅ n − 1 C r − 1 all coefficients 2 n 2^n 2 n half-sums 2 n − 1 2^{n - 1} 2 n − 1 each
If n C a = n C b {}^{n}C_{a} = {}^{n}C_{b} n C a = n C b with a ≠ b a \ne b a = b , then a + b = n a + b = n a + b = n . For a polynomial f f f : put x = 1 x = 1 x = 1 for the total and x = − 1 x = -1 x = − 1 for the alternating sum. ∑ r ⋅ n C r = n ⋅ 2 n − 1 \sum r \cdot {}^{n}C_{r} = n \cdot 2^{n - 1} ∑ r ⋅ n C r = n ⋅ 2 n − 1 . ( d + 1 ) n (d + 1)^n ( d + 1 ) n leaves remainder 1 1 1 on division by d d d , and ( d − 1 ) n (d - 1)^n ( d − 1 ) n leaves 1 1 1 or d − 1 d - 1 d − 1 . Modulo d 2 d^2 d 2 , ( d + 1 ) n ≡ 1 + n d (d + 1)^n \equiv 1 + nd ( d + 1 ) n ≡ 1 + n d . For small h h h , ( 1 + h ) n ≈ 1 + n h + n C 2 h 2 (1 + h)^n \approx 1 + nh + {}^{n}C_{2}h^2 ( 1 + h ) n ≈ 1 + nh + n C 2 h 2 . A worked example Find the sum of all the coefficients, and the sums of the even and the odd powers of x x x , in the expansion of ( 2 x + 5 ) 4 (2x + 5)^4 ( 2 x + 5 ) 4 .
Step 1. The total. Put x = 1 x = 1 x = 1 : 7 4 = 2401 7^4 = 2401 7 4 = 2401 .
Step 2. The alternating sum. Put x = − 1 x = -1 x = − 1 : 3 4 = 81 3^4 = 81 3 4 = 81 .
Step 3. The two halves. The even powers add up to 2401 + 81 2 = 1241 \dfrac{2401 + 81}{2} = 1241 2 2401 + 81 = 1241 , and the odd powers to 2401 − 81 2 = 1160 \dfrac{2401 - 81}{2} = 1160 2 2401 − 81 = 1160 .
Answer: 2401 2401 2401 , 1241 1241 1241 and 1160 1160 1160 .
Practice with new numbers
A program to try The program finds the sums of the worked example with x = 1 x = 1 x = 1 and x = − 1 x = -1 x = − 1 , and checks them against the expanded coefficients.
Run it. Do the three sums agree with the expanded coefficients?
1. Change n to 3. Predict the new total first.
Mistakes to avoid Expanding first. To add coefficients, put x = 1 x = 1 x = 1 . A half-sum of 2 n 2^n 2 n . Each half is 2 n − 1 2^{n - 1} 2 n − 1 . A negative remainder. Add the divisor to make it positive. A large h h h . A short sum works only when h h h is small. Upside-down ratio. The next coefficient is the one before times n − r + 1 r \dfrac{n - r + 1}{r} r n − r + 1 . Check yourself Find 16 C 14 {}^{16}C_{14} 16 C 14 . Stuck? Show a hint Show the answerFind the sum of all the coefficients in ( 5 x − 2 ) 3 (5x - 2)^{3} ( 5 x − 2 ) 3 . Stuck? Show a hint Show the answerFind the remainder when 8 30 8^{30} 8 30 is divided by 7 7 7 . Stuck? Show a hint Show the answerEstimate 1.03 6 1.03^{6} 1.0 3 6 with two terms. Stuck? Show a hint Show the answerIf n C 3 = n C 9 {}^{n}C_{3} = {}^{n}C_{9} n C 3 = n C 9 , find n n n . Stuck? Show a hint Show the answer