Modified pyramids

Sum of $n$ odd integers will be $n^2$ . Therefore, sum of 100 odd integers would be $100^2$ and 200 odd integers $200^2$ .

In which case $b_{100}$ can be written as $200^2 - 100^2$ while $a_{100}$ can be written as $100^2$ .

$b_{100} \div a_{100} = \frac{200^2-100^2}{100^2}$ $\boxed3$

Actually, we can generalize this for $b_n \div a_n$ .

• In the first pyramid, The sum of first $n$ odd numbers is: $\displaystyle\sum_{k=1}^{n} (2k-1)$ gives $\boxed{\boxed{n^2}}$

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• In the second pyramid, The first term of the $n^{th}$ row is given by $a+(n-1)d$ . Here $a=3$ , $d=2$ so the first term is: $3+(n-1)2 \\ =3+2n-2 \\ =\boxed{2n+1}$

• Similarly, the last term is: $3+(n-1)4 \\ =3+4n-4 \\ =\boxed{4n-1}$

• Now, the sum of all terms in the $n^{th}$ row is given by $\dfrac{n(a_1+a_n)}{2} \\ =\dfrac{n(2n+1+4n-1)}{2} \\ =\dfrac{6n^2}{2} \\ = \boxed{\boxed{3n^2}}$

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• Now $b_n \div a_n$ , $=\dfrac{3n^2}{n^2} \\=\boxed{\boxed{\boxed{3}}}$

If you take out the 100, this is basically b/a. For this I just took the first row 3 / 1 = 3, checked with the second row which is 12/4 = 3, and checked the 3rd row which is 27/9 = 3. I can assume that the pattern will be followed and thus came up with the answer 3.

same analysis, yet you made it more clearer

java code Solution using dynamic programming

public static void main(String[] args) throws IOException {

    final Integer integer = 100;
    BigInteger[] dp = new BigInteger[integer];
    dp[0] = BigInteger.ONE;
    for (int i = 1, j = 3; i < dp.length; i++, j += 2) {
        dp[i] = dp[i - 1].add(BigInteger.valueOf(j));
    }
    BigInteger[] dp2 = new BigInteger[integer];
    dp2[0] = BigInteger.valueOf(3);
    long start = 5L;
    for (int i = 1; i < dp2.length; i++, start += 2L) {
        long newStart = start;
        BigInteger integer2 = new BigInteger("0");
        for (int j = 1; j <= i + 1; j++, newStart += 2L) {
            integer2 = integer2.add(BigInteger.valueOf(newStart));
        }
        dp2[i] = integer2;
    }
    System.out.println(dp2[dp2.length - 1].divide(dp[dp.length - 1]));
}

I just noticed that the sum for pyramid 'a' was (n)^2 while the sum for pyramid 'b' was (2n)^2 -(n)^2 --------substituting 100 for each and dividing, the answer is clearly 3

We need to know the value of $b_{100}\div a_{100}$ , we note that in pyramid $b$ the difference between the first terms of two consecutive levels is $2$ , and the difference between the last terms is $4$ . Then, the the first term of its 100th row:

$3+(100-1)\times 2=201$

The last term:

$3+(100-1)\times 4=399$

Now that we have the first and last term we can calculate $b_{100}$ as the sum of an arithmetic progression:

$S_n=\dfrac{n(a_1+a_n)}{2}=\dfrac{100(201+399)}{2}=\boxed{30000}$

For $a_{100}$ we can do the same, by applying the formula:

$S_n=\dfrac{n(2a_1+(n-1)\times d)}{2}=\dfrac{100(2+(100-1)\times 2)}{2}=\boxed{10000}$

Or we can use the well known fact that the sum of the first $n$ odd numbers is $n^2$ :

$n^2=100^2=\boxed{10000}$

Finally:

$b_{100}\div a_{100}=\dfrac{30000}{10000}=\boxed{3}$