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,

$\therefore {S_{35}} = 166705$

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:
$\sum\limits_{i = 1}^n i = \dfrac{{n\left( {n + 1} \right)}}{2}$
$\sum\limits_{i = 1}^n {{i^2}} = \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}$
$\sum\limits_{i = 1}^n {{i^3}} = \dfrac{{{n^2}{{\left( {n + 1} \right)}^2}}}{4}$