"Binomial Theorem" — Practice

Question 1

Find the coefficient of x to the power 5 in the expansion of (2x squared minus 1 over x) raised to the power 10.

Accepted Answer

The general term is C(10,r) times (2x squared) to the power 10 minus r times (minus 1 over x) to the power r. Separating the numbers from the powers gives C(10,r) times 2 to the power 10 minus r times minus one to the power r times x to the power 20 minus 3r. Setting the exponent equal to 5 gives 20 minus 3r equal to 5, so r equals 5, which is a whole number in range so the term exists. Substituting gives C(10,5) times 2 to the fifth times minus one to the fifth, which is 252 times 32 times minus one, that is minus 8064. The sign comes from the factor of minus one to the power r and would be lost by ignoring the minus in the bracket.

Question 2

Find the term independent of x in the expansion of (x minus 1 over x squared) raised to the power 9, and say which term it is.

Accepted Answer

Independent of x means the power of x is zero. The general term is C(9,r) times x to the power 9 minus r times (minus 1 over x squared) to the power r, which simplifies to C(9,r) times minus one to the power r times x to the power 9 minus 3r. Setting 9 minus 3r equal to zero gives r equal to 3. The term is therefore C(9,3) times minus one cubed, which is 84 times minus one, that is minus 84. Since r equals 3 corresponds to the term numbered r plus one, this is the fourth term, not the third.

Question 3

Find the middle term in the expansion of (2x minus 3 over x) raised to the power 8.

Accepted Answer

With the index equal to 8 there are nine terms, so there is a single middle term and it is the fifth, corresponding to r equal to 4. The term is C(8,4) times (2x) to the fourth times (minus 3 over x) to the fourth. Now C(8,4) is 70, (2x) to the fourth is 16 x to the fourth, and (minus 3 over x) to the fourth is 81 over x to the fourth, the even power making the sign positive. The powers of x cancel, leaving 70 times 16 times 81, which is 90720. The term is the constant 90720, and it is independent of x because the two parts of the bracket carry x and x inverse in balance.

Question 4

How many rational terms are there in the expansion of (root 3 plus the fourth root of 5) raised to the power 20?

Accepted Answer

The general term carries 3 to the power (20 minus r) over 2 and 5 to the power r over 4. For the term to be rational both exponents must be whole numbers simultaneously. The first requires 20 minus r to be even, so r must be even. The second requires r to be a multiple of 4. The stricter condition governs, so r must be a multiple of 4. The admissible values between 0 and 20 inclusive are 0, 4, 8, 12, 16 and 20, giving six rational terms. Checking only the first condition would wrongly give eleven.

Question 5

Find the sum of all the coefficients in the expansion of (3x minus 5y) raised to the power 7, and the sum C(9,0) + C(9,1) + ... + C(9,9).

Accepted Answer

For the first, the sum of the coefficients of a polynomial is its value when every variable is set to one. Putting x and y both equal to one gives (3 minus 5) to the power 7, which is minus 2 to the power 7, that is minus 128. No expansion is needed. For the second, put both letters equal to one in the theorem with index 9. The left side becomes 2 to the power 9 and the right side is exactly the required sum, so the answer is 512. The counting reading confirms it: each of nine brackets independently contributes one of two things.

Question 6

Prove that C(n,1) + 2C(n,2) + 3C(n,3) + ... + nC(n,n) equals n times 2 to the power n minus 1.

Accepted Answer

Start from the expansion of (1 plus x) to the power n as the sum of C(n,r) x to the power r. Differentiate both sides with respect to x. The left side becomes n times (1 plus x) to the power n minus 1. The right side becomes the sum of r times C(n,r) times x to the power r minus 1, and the r equals zero term vanishes since it was constant. Now substitute x equal to 1. The left side becomes n times 2 to the power n minus 1, and the right side becomes the sum of r times C(n,r) from r equals 1 to n, which is exactly the required expression. Differentiation is what brings the factor of r down onto each coefficient.

Question 7

Evaluate (root 2 plus 1) to the power 6 plus (root 2 minus 1) to the power 6, and explain why the answer must be an integer.

Accepted Answer

The two expansions differ only in the signs of the terms carrying odd powers of 1, so adding them cancels every odd term and doubles every even one. What survives is twice the sum of C(6,0) times root 2 to the sixth, C(6,2) times root 2 to the fourth, C(6,4) times root 2 squared, and C(6,6). Those values are 8, 15 times 4 which is 60, 15 times 2 which is 30, and 1. The sum is 99, and doubling gives 198. The answer must be an integer because the surviving terms all carry even powers of root 2, which are whole numbers, and this is guaranteed by the structure of the cancellation rather than being a coincidence.

Question 8

Using a product of two expansions, prove that C(m+n, r) equals the sum over k of C(m,k) times C(n, r minus k).

Accepted Answer

Consider the identity that (1 plus x) to the power m times (1 plus x) to the power n equals (1 plus x) to the power m plus n, which is true because the indices simply add. Expand each side and compare the coefficient of x to the power r. On the right the coefficient is C(m+n, r) directly. On the left, a term in x to the power r arises by taking x to the power k from the first expansion and x to the power r minus k from the second, for every admissible k, so the coefficient is the sum over k of C(m,k) times C(n, r minus k). Equating the two gives the identity. In counting language: choosing r objects from a pool of m plus n means choosing some from the first m and the rest from the other n, then summing over how the split fell.

Question 9

Find the remainder when 7 raised to the power 103 is divided by 25.

Accepted Answer

Rewrite the base so that a multiple of 25 sits inside a bracket. Since 7 squared is 49, which is 50 minus 1, pull out one factor of 7 to make the exponent even: 7 to the power 103 equals 7 times (7 squared) to the power 51, which is 7 times (50 minus 1) to the power 51. Expanding, every term containing a factor of 50 is divisible by 25, so only the last term survives, namely minus one to the power 51, which is minus 1. So the bracket leaves a remainder of minus 1, and 7 to the power 103 leaves 7 times minus 1, which is minus 7. Adding 25 converts this to the proper remainder 18. Checking independently, 7 squared leaves minus 1 modulo 25 so the powers repeat with period four, and 103 leaves 3 on division by 4, giving 343 which leaves 18.

Question 10

Find the last two digits of 3 raised to the power 100.

Accepted Answer

The last two digits are the remainder on division by 100, so aim to get 10 inside a bracket. Write 3 to the power 100 as 9 to the power 50, which is (10 minus 1) to the power 50. Expanding, any term containing 10 squared or a higher power is divisible by 100 and can be discarded. Only two terms remain: the constant term minus one to the power 50, which is 1, and the linear term C(50,1) times 10 times minus one to the power 49, which is minus 500. But minus 500 is itself a multiple of 100 and so contributes nothing to the remainder. Therefore 3 to the power 100 leaves a remainder of 1, and its last two digits are 01.

Question 11

Show that 3 raised to the power 2n minus 1 is divisible by 8 for every positive integer n.

Accepted Answer

Rewrite the power so that 8 appears inside a bracket. Since 3 squared is 9, we have 3 to the power 2n equal to 9 to the power n, which is (8 plus 1) to the power n. Expanding by the binomial theorem, every term except the last carries at least one factor of 8, and the last term is 1 to the power n, which is 1. So 3 to the power 2n equals a multiple of 8 plus 1. Subtracting 1 leaves exactly the multiple of 8, so 3 to the power 2n minus 1 is divisible by 8 for every positive integer n. The whole proof is choosing to write 9 as 8 plus 1.

Question 12

Find the greatest term in the expansion of (1 plus x) to the power 10 when x equals two thirds, and compare its position with that of the greatest coefficient.

Accepted Answer

Compare consecutive terms rather than looking at coefficients alone, because the powers of x shrink as r grows and pull the other way. The ratio of the (r plus 1)th term to the rth term is (11 minus r) over r, times x. Substituting x equal to two thirds and requiring the ratio to exceed one gives 22 minus 2r greater than 3r, so r is less than 4.4. The terms therefore increase up to r equal to 4 and decrease afterwards, making the fifth term the greatest. It equals C(10,4) times two thirds to the fourth, which is 210 times 16 over 81, that is 1120 over 27, approximately 41.5. The greatest coefficient is C(10,5) which is 252, sitting at r equal to 5, one position further along. The two questions genuinely have different answers.

Practice — "Binomial Theorem"

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