Mo reads a lot of books. When he was five years old he read 4 books. Each year he reads three more than twice the number of books he read the previous year.

(hint: When finding the answers to (a) and (b), don't simplify too much).

(d) Use your formula from (c) to calculate how many books Mo reads when he's fifteen years old.

Check your answer by computing the same number without using the formula.

Answer

For *n* ≥ 5, let

*a*_{n} be the number of books Mo reads when he's *n* years old. We're told

*a*_{5} = 4

to start off, and that to get from one number to the next we need to multiply by 2 and then add 3.

This means for *n* > 5,

*a*_{n + 1} = 2*a*_{n} + 3.

(a) Take *a*_{5}, multiply by 2, then add 3:

*a*_{6} = 2*a*_{5} + 3

= 2(4) + 3

= 11.

Mo reads 11 books when he's 6 years old.

(b) When Mo is seven he reads

*a*_{7} = 2(11) + 3 = 25 books.

When he's eight he reads

*a*_{8} = 2(25) + 3 = 53 books.

(c) Since we're asked to give a formula for the number of books Mo reads when he's 5 + *i* years old, we'll write *a*_{6} as *a*_{5 + 1},

*a*_{7} as *a*_{5 + 2}, and so on.

Following the hint, look at how we got the answers to (a) and (b) without simplifying so much.

The first one was simple:

*a*_{5 + 1} = 2(4) + 3

Instead of simplifying, put this answer straight into the equation for *a*_{7} = *a*_{5 + 2}:

*a*_{5 + 2} = 2*a*_{5 + 1} + 3

= 2(2(4) + 3) + 3

= 2^{2}(4) + 2 × 3 + 3

Now we can put the non-simplified expression

*a*_{5 + 2} = 2^{2}(4) + 2 × 3 + 3

into the equation for *a*_{5 + 3}:

*a*_{5 + 3} = 2*a*_{5 + 2} + 3

= 2(2^{2}(4) + 2 × 3 + 3)

= 2^{3}(4) + 2^{2} × 3 + 2 × 3 + 3

Writing all our expressions next to each other, we have

*a*_{5 + 0} = 4

*a*_{5 + 1} = 2(4) + 3

*a*_{5 + 2} = 2^{2}(4) + 2 × 3 + 3

*a*_{5 + 3} = 2^{3}(4) + 2^{2} × 3 + 2 × 3 + 3

We can generalize to get an expression for *a*_{5 + i}:

*a*_{5 + i} = 2^{i}(4) + 2^{i – 1} × 3 + 2^{i – 2} × 3 + ... 2 × 3 + 3

This expression has 2 parts. The first part is 4 multiplied by a power of 2. The second part is itself a series. Writing the series in summation notation, we can rewrite the expression as

Don't be confused by the use of *j* in the summation. We need some letter for the index of summation, and *n* and *i* have already been used in this problem.

(d) We want to find

*a*_{15} =* a*_{5 + 10}.

Using the formula from (c),

The second piece of the expression is a geometric series with first term

*a* = 2^{0} × 3 = 3

and ratio 2. There are 10 terms from *j* = 0 to *j* = 9, so take *n* = 10. This partial sum is

Adding up the pieces of the expression,

*a*_{15} = 4096 + 3069 = 7165.

Check the answer by direct computation. We left off before at *a*_{8} = 53, so

*a*_{9} = 2(53) + 3 = 109

*a*_{10} = 2(109) + 3 = 221

*a*_{11} = 2(221) + 3 = 445

*a*_{12} = 2(445) + 3 = 893

*a*_{13} = 2(893) + 3 = 1789

*a*_{14} = 2(1789) + 3 = 3581

*a*_{15} = 2(3581) + 3 = 7165

Thankfully, this answer agrees with what we got using the formula. We conclude that Mo reads an absolutely ludicrous number of books the year he's 15: 7165 books.