For any natural number k,leta_k = 3^k.Thesmallestnaturalnumbermforwhich\{(a_1)^1 × (a_2)^2 × \ldots × (a_{20})^{20}\} < \{a_{21} × a_{22} × \ldots × a_{(20+m)}\} , is
⌨️ Shortcuts:[1-4 / A-D] Select[Enter] Check[B] Star[S] Solution
Official Correct Answer: C. "For any natural number \( k \), let \( a_k = 3^k \). The smallest natural number \( m \) for which \( \{(a_1)^1 \times (a_2)^2 \times \ldots \times (a_{20})^{20}\} < \{a_{21} \times a_{22} \times \ldots \times a_{(20+m)}\} \), is...?"
And the options are A) 59, B) 56, C) 58, D) 57.
Hmm, okay. So, We find the smallest \( m \) such that the product on the left is less than the product on the right.
First, let me parse the expressions. The left side is \( (a_1)^1 \times (a_2)^2 \times \ldots \times (a_{20})^{20} \). Since \( a_k = 3^k \), each term \( a_i \) is \( 3^i \). So, substituting that in, the left side becomes:
\( (3^1)^1 \times (3^2)^2 \times \ldots \times (3^{20})^{20} \).
Similarly, the right side is \( a_{21} \times a_{22} \times \ldots \times a_{20 + m} \), which is \( 3^{21} \times 3^{22} \times \ldots \times 3^{20 + m} \).
I think I can simplify both sides by using properties of exponents. Let me start with the left side.
Left Side:
Each term is \( (3^k)^k = 3^{k^2} \). So, the entire product is \( 3^{1^2} \times 3^{2^2} \times \ldots \times 3^{20^2} \). When you multiply exponents with the same base, you add the exponents. So, the total exponent is \( 1^2 + 2^2 + 3^2 + \ldots + 20^2 \).
I remember the formula for the sum of squares up to \( n \) is \( \frac{n(n + 1)(2n + 1)}{6} \). Compute that for \( n = 20 \).
Sum of squares from 1 to 20:
\( S = \frac{20 \times 21 \times 41}{6} \).
Calculating step by step:
20 divided by 6 is \( \frac{10}{3} \), but maybe it's better to compute numerator first:
20 × 21 = 420
420 × 41 = Computing 420 × 40 = 16,800 and 420 × 1 = 420, so total is 17,220.
Then, 17,220 divided by 6 is 2,870. So, the exponent on the left side is 2,870. Therefore, left side is \( 3^{2870} \).
Right Side:
The product is \( 3^{21} \times 3^{22} \times \ldots \times 3^{20 + m} \). Again, when multiplying exponents with the same base, we add the exponents. So, the total exponent is the sum from 21 to (20 + m).
Wait, the terms go from 21 to (20 + m). Let \( n = 20 + m \). So, the sum is from 21 to \( n \).
The sum of consecutive integers from \( a \) to \( b \) is \( \frac{(b + a)(b - a + 1)}{2} \). Here, \( a = 21 \), \( b = n \). So, the sum is \( \frac{(n + 21)(n - 20)}{2} \).
But \( n = 20 + m \), so substituting:
Sum = \( \frac{(20 + m + 21)(20 + m - 20)}{2} \) = \( \frac{(41 + m)(m)}{2} \).
So, the exponent on the right side is \( \frac{m(m + 41)}{2} \). Therefore, the right side is \( 3^{\frac{m(m + 41)}{2}} \).
Now, the inequality given is:
\( 3^{2870} < 3^{\frac{m(m + 41)}{2}} \).
Since the base 3 is the same and greater than 1, the inequality holds if the exponent on the left is less than the exponent on the right. So, we can write:
\( 2870 < \frac{m(m + 41)}{2} \).
Multiply both sides by 2:
\( 5740 < m(m + 41) \).
So, we have the quadratic inequality:
\( m^2 + 41m - 5740 > 0 \).
We need to solve for \( m \). Let's write the quadratic equation:
\( m^2 + 41m - 5740 = 0 \).
To find the roots, we can use the quadratic formula:
\( m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).
Here, \( a = 1 \), \( b = 41 \), \( c = -5740 \).
Compute discriminant \( D = b^2 - 4ac = 41^2 - 4(1)(-5740) \).
41 squared is 1681. Then, 4 × 5740 = 22,960. So, D = 1681 + 22,960 = 24,641.
Square root of 24,641: , 150 squared is 22,500, 160 squared is 25,600. So, it's between 150 and 160. Let me try 157: 157^2 = 24,649. Hmm, that's 8 more than 24,641. So, 157^2 is 24,649, which is