Proof by Induction (Edexcel International A Level (IAL) Further Maths: Further Pure 1): Exam Questions

Exam code: YFM01

2 hours13 questions
1a
5 marks

A sequence of positive numbers is defined by

space space space space space u subscript 1 equals 5
u subscript n plus 1 end subscript equals 3 u subscript n plus 2 comma    n greater or equal than 1

Prove by induction that, for n element of straight integer numbers to the power of plus ,

u subscript n equals 2 cross times left parenthesis 3 right parenthesis to the power of n minus 1

1b
6 marks

Prove by induction that, for n element of straight integer numbers to the power of plus,

sum from r equals 1 to n of fraction numerator 4 r over denominator 3 to the power of r end fraction equals 3 minus fraction numerator left parenthesis 3 plus 2 n right parenthesis over denominator 3 to the power of n end fraction

2a
4 marks

Prove by induction that for n element of straight integer numbers to the power of plus

open parentheses table row 1 r row 0 2 end table close parentheses to the power of n equals open parentheses table row 1 cell left parenthesis 2 to the power of n minus 1 right parenthesis r end cell row 0 cell 2 to the power of n end cell end table close parentheses

where r is a constant.

2b
3 marks

bold M equals open parentheses table row 4 0 row 0 5 end table close parentheses bold N equals open parentheses table row 1 cell negative 2 end cell row 0 2 end table close parentheses to the power of 4

The transformation represented by matrix M followed by the transformation represented by matrix N is represented by the matrix B

(i) Determine N in the form open parentheses table row a b row c d end table close parentheses where a, b, c and d are integers.

[1]

(ii) Determine B

[2]

2c
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2 marks

Hexagon S is transformed onto hexagon S apostrophe by matrix B

Given that the area of S apostrophe is 720 square units, determine the area of S

3
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6 marks

Prove by induction that for n element of straight integer numbers to the power of plus

straight f left parenthesis n right parenthesis equals 7 to the power of n minus 1 end exponent plus 8 to the power of 2 n plus 1 end exponent

is divisible by 57

4a
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5 marks

Prove by induction that for n element of straight integer numbers to the power of plus

open parentheses table row 5 cell negative 1 end cell row 4 1 end table close parentheses to the power of n equals 3 to the power of n minus 1 end exponent open parentheses table row cell 2 n plus 3 end cell cell negative n end cell row cell 4 n end cell cell 3 minus 2 n end cell end table close parentheses

4b
5 marks

Prove by induction that for n element of straight integer numbers to the power of plus

straight f left parenthesis n right parenthesis equals 8 to the power of 2 n plus 1 end exponent plus 6 to the power of 2 n minus 1 end exponent

is divisible by 7

5
5 marks

Prove, by induction, that for n element of straight integer numbers, n greater-than or slanted equal to 2

4 to the power of n plus 6 n minus 10

is divisible by 18

6
6 marks

Prove by induction that for all positive integers n

sum from r equals 1 to n of log left parenthesis 2 r minus 1 right parenthesis equals log space blank open parentheses fraction numerator left parenthesis 2 n right parenthesis factorial over denominator 2 to the power of n   n factorial end fraction close parentheses

7a
5 marks

A sequence of numbers is defined by

u subscript 1 equals 3

u subscript n plus 1 end subscript equals 2 u subscript n minus 2 to the power of n plus 1 end exponent      n greater-than or slanted equal to 1

Prove by induction that, for n element of straight natural numbers

u subscript n equals 5 cross times 2 to the power of n minus 1 end exponent minus n cross times 2 to the power of n

7b
5 marks

Prove by induction that, for n element of straight natural numbers

straight f left parenthesis n right parenthesis equals 5 to the power of n plus 2 end exponent minus 4 n minus 9

is divisible by 16

8a
5 marks

Prove by induction that, for n element of straight natural numbers

sum from r equals 1 to n of r cubed equals 1 fourth n squared left parenthesis n plus 1 right parenthesis squared

8b
4 marks

Using the standard summation formulae, show that

sum from r equals 1 to n of r left parenthesis r plus 1 right parenthesis left parenthesis r minus 1 right parenthesis equals 1 fourth n left parenthesis n plus A right parenthesis left parenthesis n plus B right parenthesis left parenthesis n plus C right parenthesis

where A, B and C are constants to be determined.

8c
1 mark

Determine the value of n for which

3 sum from r equals 1 to n of r left parenthesis r plus 1 right parenthesis left parenthesis r minus 1 right parenthesis equals 17 sum from r equals n to 2 n of r squared

9a
5 marks

A sequence of numbers is defined by

u subscript 1 equals 0      u subscript 2 equals negative 6

u subscript n plus 2 end subscript equals 5 u subscript n plus 1 end subscript minus 6 u subscript n      n greater or equal than 1

Prove by induction that, for n element of straight integer numbers to the power of plus

u subscript n equals 3 cross times 2 to the power of n minus 2 cross times 3 to the power of n

9b
5 marks

Prove by induction that, for all positive integers n,

f left parenthesis n right parenthesis equals 3 to the power of 3 n minus 2 end exponent plus 2 to the power of 4 n minus 1 end exponent

is divisible by 11

10a
5 marks

Prove by induction that for n element of straight natural numbers

sum from r equals 1 to n of r squared equals n over 6 left parenthesis n plus 1 right parenthesis left parenthesis 2 n plus 1 right parenthesis

10b
4 marks

Hence show that

sum from r equals 1 to n of left parenthesis r squared plus 2 right parenthesis equals n over 6 left parenthesis a n squared plus b n plus c right parenthesis

where a, b and c are integers to be found.

10c
2 marks

Using your answers to part (b), find the value of

sum from r equals 10 to 25 of left parenthesis r squared plus 2 right parenthesis

11
6 marks

Prove by induction that 4 to the power of n plus 2 end exponent plus 5 to the power of 2 n plus 1 end exponent is divisible by 21 for all positive integers n.

12a
6 marks

A sequence of numbers u subscript 1, u subscript 2, u subscript 3,....is defined by

u subscript n plus 1 end subscript equals 1 third left parenthesis 2 u subscript n minus 1 right parenthesis comma      u subscript 1 equals 1

Prove by induction that, for n element of straight integer numbers to the power of plus

u subscript n equals 3 open parentheses 2 over 3 close parentheses to the power of n minus 1

12b
6 marks

f left parenthesis n right parenthesis equals 2 to the power of n plus 2 end exponent plus 3 to the power of 2 n plus 1 end exponent

Prove by induction that, for n element of straight integer numbers to the power of plus , f left parenthesis n right parenthesis is a multiple of 7.

13a
6 marks

(i) Prove by induction that, for n element of straight integer numbers to the power of plus

sum from r equals 1 to n of fraction numerator 2 r squared minus 1 over denominator r squared left parenthesis r plus 1 right parenthesis squared end fraction equals fraction numerator n squared over denominator left parenthesis n plus 1 right parenthesis squared end fraction

[6]

(ii) Prove by induction that, for n element of straight integer numbers to the power of plus

straight f left parenthesis n right parenthesis equals 12 to the power of n plus 2 cross times 5 to the power of n minus 1 end exponent

is divisible by 7.

[6]

13b
6 marks

Prove by induction that, for n element of straight integer numbers to the power of plus

straight f left parenthesis n right parenthesis equals 12 to the power of n plus 2 cross times 5 to the power of n minus 1 end exponent

is divisible by 7.