Question

A number is a good number when the number is the product of two consecutive odd numbers. Find the sum of all four-digit good numbers.

Answer 1

First find the consecutive odd numbers whose product is the smallest four-digit good number. After that find the consecutive odd numbers whose product is the largest four-digit good number. Then, find the general term of the good number. Now, find the number of terms by the arithmetic progression formula, ${a_n} = {a_1} + \left( {n - 1} \right)d$. After that use the summation property to find the sum of the good numbers. Then, substitute the value and do calculation. The outcome is the desired result.

Complete step-by-step solution:
Given: - A number is a good number when the number is the product of two consecutive odd numbers.
The consecutive odd numbers whose product is the smallest four-digit good number is $31 \times 33$.
The consecutive odd numbers whose product is the largest four-digit good number is $99 \times 101$.
So, the sum of all four-digit good number is,
$31 \times 33 + 33 \times 35 + \cdots + 99 \times 101$
The general term is,
${T_n} = \left[ {31 + 2\left( {n - 1} \right)} \right]\left[ {33 + 2\left( {n - 1} \right)} \right]$
Open the brackets,
$\Rightarrow {T_n} = \left( {31 + 2n - 2} \right)\left( {33 + 2n - 2} \right)$
Subtract the like terms,
$\Rightarrow {T_n} = \left( {29 + 2n} \right)\left( {31 + 2n} \right)$
Now, multiply the terms,
$\Rightarrow {T_n} = 899 + 62n + 58n + 4{n^2}$
Add the like terms,
$\Rightarrow {T_n} = 899 + 120n + 4{n^2}$
For number of terms, use the general term of arithmetic progression,
${a_n} = {a_1} + \left( {n - 1} \right)d$
Here, ${a_1} = 31$, ${a_n} = 99$ and $d = 2$.
Substitute these values in the formula,
$\Rightarrow 99 = 31 + \left( {n - 1} \right) \times 2$
Move 31 to the other side and subtract from 99,
$\Rightarrow 2\left( {n - 1} \right) = 68$
Divide both side by 2,
$\Rightarrow n - 1 = 34$
Move 1 to the other side and add,
$\Rightarrow n = 35$............…… (1)
So, the sum can be represented as,
${S_n} = \sum\limits_{i = 1}^n {{T_i}}$
Substitute the value of ${T_n}$,
$\Rightarrow {S_n} = \sum\limits_{i = 1}^n {\left( {899 + 120i + {i^2}} \right)}$
Then,
$\Rightarrow {S_n} = \sum\limits_{i = 1}^n {899} + \sum\limits_{i = 1}^n {120i} + \sum\limits_{i = 1}^n {4{i^2}}$
Use the summation formula to calculate the value,
$\Rightarrow {S_n} = 899n + 120 \times \dfrac{{n\left( {n + 1} \right)}}{2} + 4 \times \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}$
Substitute the value of $n$ from equation (1),
$\Rightarrow {S_{35}} = 899 \times 35 + 120 \times \dfrac{{35\left( {35 + 1} \right)}}{2} + 4 \times \dfrac{{35\left( {35 + 1} \right)\left( {2 \times 35 + 1} \right)}}{6}$
Add the terms in the brackets,
$\Rightarrow {S_{35}} = 899 \times 35 + 120 \times \dfrac{{35 \times 36}}{2} + 4 \times \dfrac{{35 \times 36 \times 71}}{6}$
Cancel out the common terms and multiply the terms,
$\Rightarrow {S_{35}} = 31465 + 75600 + 59640$
Add the terms on the right side,
$\therefore {S_{35}} = 166705$

The sum of all 4 digit good numbers is 166705.

Note: A series is defined as the sum of the terms of a sequence. It is denoted by $\sum\limits_{i = 1}^n {{a_i}}$, where ${a_i}$ is the $i^{th}$ term of the sequence and i is a variable. $\sum$ is a symbol which stands for ‘summation’.
The formulas of sum of $n,{n^2},{n^3}$ are:
$\sum\limits_{i = 1}^n i = \dfrac{{n\left( {n + 1} \right)}}{2}$
$\sum\limits_{i = 1}^n {{i^2}} = \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}$
$\sum\limits_{i = 1}^n {{i^3}} = \dfrac{{{n^2}{{\left( {n + 1} \right)}^2}}}{4}$