Question

A gardener wanted to reward a boy for his good deeds by giving some mangoes. He gave the boy two choices. He could either have $1000$ mangoes at once or he could get $1$ mango on the first day, $2$ on the second day, $4$ on the third day, $8$ mangoes on the fourth day and so on for ten days. Which option should the boy choose to get the maximum number of mangoes?

Answer 1

It is given that the boy gets $1$ mango on the first day, $2$ mangoes on the second day, $4$ on third day, $8$ on the fourth day and so on for ten days. Let us write this as sequence as follows:

$1,2,4,8,....10$ days

We need to make this sequence as series because we have to find the sum of mangoes in ten days. If the number of mangoes collected in this way is greater then $1000$ we can say that this is the better way. So the above sequence will become,

$1+2+4+8+.....10$ days

In the above series, the first term is $a_1=1$, the second term is $a_2=2$ and so on.

We find the common ratio $r$ by dividing the second term by first term as shown below:

$r=\frac{2}{1}=2>1$

We know that the sum of an geometric series with first term $a$ and common ratio $r$ is $S_n=\frac{a(r^n-1)}{r-1}$ if $r>1$

Now, substitute $a=1,r=2$ and $n=10$ in $S_n=\frac{a(r^n-1)}{r-1}$ as follows:

$S_{10}=\frac{1[{\left(2\right)}^{10}-1]}{2-1}=\frac{1[{\left(2\right)}^{10}-1]}{1}={\left(2\right)}^{10}-1=1024-1=1023$

Hence, in this way the boy can get more mangoes than the previous way. So the boy have to choose the second way to get more mangoes.