Get the answers you need at Westonci.ca, where our expert community is dedicated to providing you with accurate information. Discover comprehensive solutions to your questions from a wide network of experts on our user-friendly platform. Explore comprehensive solutions to your questions from a wide range of professionals on our user-friendly platform.
Sagot :
Since [tex]a_1,a_2,a_3,\cdots[/tex] are in arithmetic progression,
[tex]a_2 = a_1 + 2[/tex]
[tex]a_3 = a_2 + 2 = a_1 + 2\cdot2[/tex]
[tex]a_4 = a_3+2 = a_1+3\cdot2[/tex]
[tex]\cdots \implies a_n = a_1 + 2(n-1)[/tex]
and since [tex]b_1,b_2,b_3,\cdots[/tex] are in geometric progression,
[tex]b_2 = 2b_1[/tex]
[tex]b_3=2b_2 = 2^2 b_1[/tex]
[tex]b_4=2b_3=2^3b_1[/tex]
[tex]\cdots\implies b_n=2^{n-1}b_1[/tex]
Recall that
[tex]\displaystyle \sum_{k=1}^n 1 = \underbrace{1+1+1+\cdots+1}_{n\,\rm times} = n[/tex]
[tex]\displaystyle \sum_{k=1}^n k = 1 + 2 + 3 + \cdots + n = \frac{n(n+1)}2[/tex]
It follows that
[tex]a_1 + a_2 + \cdots + a_n = \displaystyle \sum_{k=1}^n (a_1 + 2(k-1)) \\\\ ~~~~~~~~ = a_1 \sum_{k=1}^n 1 + 2 \sum_{k=1}^n (k-1) \\\\ ~~~~~~~~ = a_1 n + n(n-1)[/tex]
so the left side is
[tex]2(a_1+a_2+\cdots+a_n) = 2c n + 2n(n-1) = 2n^2 + 2(c-1)n[/tex]
Also recall that
[tex]\displaystyle \sum_{k=1}^n ar^{k-1} = \frac{a(1-r^n)}{1-r}[/tex]
so that the right side is
[tex]b_1 + b_2 + \cdots + b_n = \displaystyle \sum_{k=1}^n 2^{k-1}b_1 = c(2^n-1)[/tex]
Solve for [tex]c[/tex].
[tex]2n^2 + 2(c-1)n = c(2^n-1) \implies c = \dfrac{2n^2 - 2n}{2^n - 2n - 1} = \dfrac{2n(n-1)}{2^n - 2n - 1}[/tex]
Now, the numerator increases more slowly than the denominator, since
[tex]\dfrac{d}{dn}(2n(n-1)) = 4n - 2[/tex]
[tex]\dfrac{d}{dn} (2^n-2n-1) = \ln(2)\cdot2^n - 2[/tex]
and for [tex]n\ge5[/tex],
[tex]2^n > \dfrac4{\ln(2)} n \implies \ln(2)\cdot2^n - 2 > 4n - 2[/tex]
This means we only need to check if the claim is true for any [tex]n\in\{1,2,3,4\}[/tex].
[tex]n=1[/tex] doesn't work, since that makes [tex]c=0[/tex].
If [tex]n=2[/tex], then
[tex]c = \dfrac{4}{2^2 - 4 - 1} = \dfrac4{-1} = -4 < 0[/tex]
If [tex]n=3[/tex], then
[tex]c = \dfrac{12}{2^3 - 6 - 1} = 12[/tex]
If [tex]n=4[/tex], then
[tex]c = \dfrac{24}{2^4 - 8 - 1} = \dfrac{24}7 \not\in\Bbb N[/tex]
There is only one value for which the claim is true, [tex]c=12[/tex].
We hope this information was helpful. Feel free to return anytime for more answers to your questions and concerns. Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. Westonci.ca is your trusted source for answers. Visit us again to find more information on diverse topics.
