Algebraic Proof

Check whether each bracket is always odd, and how far apart the two brackets are.

1. 2n is always even, so 2n+1 is always odd
2. The next odd number is 2 more: 2n+3
3. n(n+1) is two consecutive integers; 2n+2 is even; 2n-1 and 2n+3 are 4 apart
4. Answer: (2n+1)(2n+3)

A proof must work for every value of n: look for a statement that writes the expression as 6 times an integer.

1. To show a number is a multiple of 6, write it as 6\times(an integer)
2. 12n+18=6(2n+3) and 2n+3 is an integer when n is
3. One value of n proves nothing about all n; a factor of 2 only shows it is even; two even numbers can add to a number that is not a multiple of 6 (e.g. 12+14=26)
4. Answer: 12n+18=6(2n+3), and 2n+3 is an integer, so 12n+18 is a multiple of 6

Try factorising the terms involving n and think about whether the product is odd or even when n is odd and when n is even.

1. n^2+3n=n(n+3)
2. If n is even, n(n+3) is even
3. If n is odd, n+3 is even, so n(n+3) is even
4. So n^2+3n is always even, and n^2+3n+5 is always odd
5. The others are even when n is odd (e.g. n=1 gives 6, 8 and 8)
6. Answer: n^2+3n+5

Write down all five integers in terms of n, add them, then take out the common factor.

1. The integers are n, n+1, n+2, n+3, n+4
2. Sum =5n+10
3. =5(n+2), so the sum of five consecutive integers is always a multiple of 5
4. Answer: 5(n+2)

Expand each part separately, then put the second expansion in a bracket before you subtract.

1. (n+5)^2=n^2+10n+25
2. (n+1)(n+9)=n^2+10n+9
3. n^2+10n+25-(n^2+10n+9)
4. =n^2+10n+25-n^2-10n-9
5. Answer: 16

Expand the brackets, then take out a factor of 2 from every term except the 1.

1. (2m+1)(2n+1)=4mn+2m+2n+1
2. =2(2mn+m+n)+1
3. 2mn+m+n is an integer, so the product of two odd numbers is always odd
4. Answer: k=2mn+m+n

Expand each product separately, then put the second expansion in a bracket before you subtract.

1. (5k+2)(k+3)=5k^2+15k+2k+6=5k^2+17k+6
2. (5k-1)(k+2)=5k^2+10k-k-2=5k^2+9k-2
3. Subtract, keeping the bracket: 5k^2+17k+6-(5k^2+9k-2)
4. =5k^2+17k+6-5k^2-9k+2
5. =8k+8
6. =8(k+1), which is a multiple of 8
7. Answer: 8k+8

Check what happens to every term inside the bracket when it is subtracted.

1. Line 1 is right: (4k+3)(k+2)=4k^2+8k+3k+6
2. Subtracting the bracket changes every sign: -4k^2-11k-6
3. Line 2 has +6, so Line 2 is the first error
4. (The correct result is 9k-6=3(3k-2), so the statement is true but this proof is not)
5. Answer: Line 2

Look for a value of n that makes every term of n^2-n+11 share a common factor.

1. n=1,2,\ldots,10 give 11, 13, 17, 23, 31, 41, 53, 67, 83, 101, which are all prime
2. n=11: 11^2-11+11=121
3. 121=11\times11, which is not prime
4. Answer: n=11

Write the result as an expression and take out the common factor (n+1).

1. n(n+1)+(n+1)=(n+1)(n+1)
2. =(n+1)^2, which is the square of an integer
3. n=1 gives 4: not odd, not a multiple of 3 and not prime
4. Answer: it is a square number

Expand both squares and simplify the left-hand side, then compare it with 24n.

1. (n+a)^2=n^2+2an+a^2
2. (n-a)^2=n^2-2an+a^2
3. Subtract: n^2+2an+a^2-(n^2-2an+a^2)=4an
4. 4an\equiv24n, so 4a=24
5. Answer: a=6

Complete the square and think about the smallest value the squared bracket can take.

1. x^2+8x+k=(x+4)^2-16+k
2. (x+4)^2\geqslant0, and it equals 0 when x=-4
3. So the smallest value of the expression is k-16
4. For it to be always positive, k-16>0, so k>16
5. Answer: k=17

Replace n with (n+1) to get the (n+1)th term, and keep it in brackets.

1. (n+1)th term: 3(n+1)^2+(n+1)
2. =3n^2+6n+3+n+1=3n^2+7n+4
3. Subtract: 3n^2+7n+4-(3n^2+n)
4. =6n+4=2(3n+2), so the difference between consecutive terms is always even
5. Answer: 6n+4

Put the second expansion in a bracket before you subtract, so every sign inside it changes.

1. (3k+5)(k+3)=3k^2+14k+15
2. (3k-2)(k-2)=3k^2-8k+4
3. 3k^2+14k+15-(3k^2-8k+4)
4. =22k+11
5. Answer: 11(2k+1), an odd multiple of 11

Expand each cube as (n+2)(n+2)^2, and bracket the second expansion before subtracting.

1. (n+2)^3=n^3+6n^2+12n+8
2. (n-2)^3=n^3-6n^2+12n-8
3. n^3+6n^2+12n+8-(n^3-6n^2+12n-8)
4. =12n^2+16
5. Answer: 4(3n^2+4), so it is always a multiple of 4

Take out the factor 3 from the x^2 and x terms first. Answer part (a) before attempting part (b).

Part (a)

1. Take out the factor 3 from the x terms: 3(x^2+6x)+31
2. x^2+6x=(x+3)^2-9
3. 3\left((x+3)^2-9\right)+31=3(x+3)^2-27+31
4. Answer: 3(x+3)^2+4

Part (b)

1. A square is never negative: (x+3)^2\geqslant0
2. So 3(x+3)^2\geqslant0 and 3(x+3)^2+4\geqslant4>0
3. One value of x proves nothing; the second statement ignores the sign of (x+3)^2; (x+3)^2=0 when x=-3, so it is not always >0
4. Answer: (x+3)^2\geqslant0, so 3(x+3)^2+4\geqslant4, which is greater than 0

Treat 2n^2+2n+1 and 2n^2+2n as A and B and use A^2-B^2=(A-B)(A+B). Answer part (a) before attempting part (b).

Part (a)

1. Use the difference of two squares, A^2-B^2=(A-B)(A+B)
2. A-B=(2n^2+2n+1)-(2n^2+2n)=1
3. A+B=4n^2+4n+1
4. So the result is 4n^2+4n+1=(2n+1)^2, which proves the sides fit Pythagoras' theorem
5. Answer: 4n^2+4n+1

Part (b)

1. The shortest side is 2n+1, so 2n+1=21 and n=10
2. Hypotenuse =2n^2+2n+1=200+20+1
3. Answer: 221 cm

Expand (n+3)^2 first, then multiply by (n+3). Answer part (a) before attempting part (b).

Part (a)

1. (n+3)^2=n^2+6n+9
2. (n+3)(n^2+6n+9)=n^3+6n^2+9n+3n^2+18n+27
3. Answer: n^3+9n^2+27n+27

Part (b)

1. (n+3)^3-n^3-27=n^3+9n^2+27n+27-n^3-27
2. =9n^2+27n
3. Answer: 9n(n+3)

Part (c)

1. 9n(n+3) is a multiple of 18 when n(n+3) is even
2. n even: n(n+3) is even. n odd: n+3 is even, so n(n+3) is even
3. So n(n+3)=2p and 9n(n+3)=18p
4. (n and n+3 are not consecutive; a multiple of 9 need not be a multiple of 18; one value proves nothing)
5. Answer: the second statement

Bracket the second expansion before subtracting so every sign changes, then think about what n+1 is when n is odd.

Part (a)

1. (2n+7)^2=4n^2+28n+49
2. (2n-3)^2=4n^2-12n+9
3. Subtract, keeping the bracket: 4n^2+28n+49-(4n^2-12n+9)
4. =40n+40
5. =40(n+1)
6. (Or use the difference of two squares: (2n+7-2n+3)(2n+7+2n-3)=10(4n+4)=40(n+1))

Part (b)

1. If n is odd, n+1 is even
2. So n+1=2p for some integer p
3. 40(n+1)=40\times 2p=80p, a multiple of 80
4. n=1 gives 80 and n=3 gives 160, so no larger number always works
5. m=80

Use the second differences to find the coefficient of n^2. Answer part (a) before attempting part (b).

Part (a)

1. First differences: 6, 8, 10, 12; second difference: 2, so the nth term starts n^2
2. Sequence minus n^2: 3, 6, 9, 12, 15, which is 3n
3. Answer: n^2+3n

Part (b)

1. 4T+9=4(n^2+3n)+9
2. =4n^2+12n+9
3. (an+b)^2=a^2n^2+2abn+b^2, so a^2=4 and b^2=9: a=2, b=3 (check: 2ab=12)
4. Answer: (2n+3)^2

Part (c)

1. If T is a term, 4T+9 must be a square number
2. 4\times180+9=729=27^2, 4\times270+9=1089=33^2, 4\times340+9=1369=37^2
3. 4\times300+9=1209, which is not a square number (34^2=1156, 35^2=1225)
4. Answer: 300