Let u_n be the sequence defined as u_n=3\times 2^n.
Calculate \sum_{k=2}^7u_k.
The sequence u_n = 3 \times 2^{n} is a geometric sequence. Therefore, a sum of consecutive terms in this sequence can be calculated with the following formula:
S_n = \dfrac{a\left(r^n - 1\right)}{r - 1}
Where a is the first term in the sum, r is the common ratio between terms, and n is the total number of terms to be summed. Here:
- r = 2
- a = u_2 = 3 \times 2^2 = 3 \times 4 = 12
- n = 6
Therefore:
\sum_{k=2}^{7} u_k = \dfrac{12\left(2^6 - 1\right)}{2 - 1} = 12\left(64 - 1\right) = 12\left(63\right) = 756
\sum_{k=2}^{7} u_k = 756
Let u_n be the sequence defined as u_n = \dfrac{2}{2^n}.
Calculate \sum_{k=2}^{8} u_k.
The sequence u_n = \dfrac{2}{2^n} is a geometric sequence. To find this ratio, realize that the expression for the sequence can be re-written as:
u_n={2\left(\dfrac{1}{2}\right)^{n}}
Therefore, a sum of consecutive terms in this sequence can be calculated with the following formula:
S_n = \dfrac{a\left(r^n - 1\right)}{r - 1}
Where a is the first term in the sum, r is the common ratio between terms, and n is the total number of terms to be summed. Here:
- r =\dfrac{1}{2}
- a = u_2 = \dfrac{2}{2^2} = \dfrac{2}{4} = \dfrac{1}{2}
- n = 7
Therefore:
\sum_{k=2}^{8} u_k = \dfrac{\dfrac{1}{2}\left(\left(\dfrac{1}{2}\right)^7 - 1\right)}{\dfrac{1}{2} - 1}= \dfrac{\dfrac{1}{2}\left(\dfrac{1}{128} - 1\right)}{\dfrac{-1}{2}}= \dfrac{127}{128}
\sum_{k=2}^{8} u_k = \dfrac{127}{128}
Let u_n be the sequence defined as u_n = \dfrac{1}{3} \times \left(\dfrac{5}{4}\right)^n.
Calculate \sum_{k=1}^{4} u_k.
The sequence u_n = \dfrac{1}{3} \times \left(\dfrac{5}{4}\right)^n is a geometric sequence. Therefore, a sum of consecutive terms in this sequence can be calculated with the following formula:
S_n = \dfrac{a\left(r^n - 1\right)}{r - 1}
Where a is the first term in the sum, r is the common ratio between terms, and n is the total number of terms to be summed. Here:
- r = \dfrac{5}{4}
- a = u_1 = \dfrac{1}{3} \times \left(\dfrac{5}{4}\right)^1 = \dfrac{5}{12}
- n = 4
Therefore:
\sum_{k=1}^{4} u_k = \dfrac{\dfrac{5}{12}\left(\left(\dfrac{5}{4}\right)^4 - 1\right)}{\dfrac{5}{4} - 1}= \dfrac{\dfrac{5}{12}\left(\dfrac{625}{256} - 1\right)}{\dfrac{1}{4}}= \dfrac{5}{3}\left(\dfrac{369}{256}\right)= \dfrac{615}{256}
\sum_{k=1}^{4} u_k = \dfrac{615}{256}
Let u_n be the sequence defined as u_n = 6 \times 4^n.
Calculate \sum_{k=4}^{7} u_k.
The sequence u_n = 6 \times 4^n is a geometric sequence. Therefore, a sum of consecutive terms in this sequence can be calculated with following the formula:
S_n = \dfrac{a\left(r^n - 1\right)}{r - 1}
Where a is the first term in the sum, r is the common ratio between terms, and n is the total number of terms to be summed. Here:
- r = 4
- a = u_4 = 6 \times 4^4 = 6 \times 256 = 1\ 536
- n = 4
Therefore:
\sum_{k=4}^{7} u_k = \dfrac{1\ 536\left(4^4 - 1\right)}{4 - 1}= \dfrac{1\ 536\left(256 - 1\right)}{3}= 512\left(255\right)= 130\ 560
\sum_{k=4}^{7} u_k = 130\ 560
Let u_n be the sequence defined as u_n = 3 \times \left(\dfrac{2}{3}\right)^n.
Calculate \sum_{k=3}^{9} u_k.
The sequence u_n = 3 \times \left(\dfrac{2}{3}\right)^n is a geometric sequence. Therefore, a sum of consecutive terms in this sequence can be calculated with the following formula:
S_n = \dfrac{a\left(r^n - 1\right)}{r - 1}
Where a is the first term in the sum, r is the common ratio between terms, and n is the total number of terms to be summed. Here:
- r = \dfrac{2}{3}
- a = u_3 = 3 \times \left(\dfrac{2}{3}\right)^3 = 3 \times \left(\dfrac{8}{27}\right) = \dfrac{8}{9}
- n = 7
Therefore:
\sum_{k=3}^{9} u_k = \dfrac{\dfrac{8}{9}\left(\left(\dfrac{2}{3}\right)^7 - 1\right)}{\dfrac{2}{3} - 1}= \dfrac{\dfrac{8}{9}\left(\dfrac{128}{2\ 187} - 1\right)}{\dfrac{-1}{3}}= \dfrac{-8}{3}\left(\dfrac{-2\ 059}{2\ 187}\right)= \dfrac{16\ 472}{6\ 561}
\sum_{k=3}^{9} u_k=\dfrac{16\ 472}{6\ 561}
Let u_n be the sequence defined as u_n = 5 \times 2^{2n}.
Calculate \sum_{k=1}^{6} u_k.
The sequence u_n = 5 \times 2^{2n} is a geometric sequence. To find out the common ratio, notice that since 2^{2n} = 2^{2^n} = 4^n, the sequence can be written as:
u_n = 5 \times 4^{n}
Therefore, a sum of consecutive terms in this sequence can be calculated with the following formula:
S_n = \dfrac{a\left(r^n - 1\right)}{r - 1}
Where a is the first term in the sum, r is the common ratio between terms, and n is the total number of terms to be summed. Here:
- r = 4
- a = u_1 = 5 \times 2^{2\left(1\right)} = 5 \times 4 = 20
- n = 6
Therefore:
\sum_{k=1}^{6} u_k = \dfrac{20\left(4^6 - 1\right)}{4 - 1}= \dfrac{20\left(4\ 096 - 1\right)}{3}= \dfrac{20\left(4\ 095\right)}{3}= 20\left(1\ 365\right)= 27\ 300
\sum_{k=1}^{6} u_k = 27\ 300
Let u_n be the sequence defined as u_n = \left(\dfrac{2}{3^n}\right)^2.
Calculate \sum_{k=2}^{5} u_k.
The sequence u_n = \left(\dfrac{2}{3^n}\right)^2 is a geometric sequence. To recognize this and find the common ratio, rewrite the sequence :
u_n =\left(\dfrac{2}{3^n}\right)^2=\dfrac{2^2}{3^{2n}}=4 \times \dfrac{1}{\left(3^2\right)^n}=4 \times \dfrac{1}{9^n}=4 \times \left(\dfrac{1}{9}\right)^n
Therefore, a sum of consecutive terms in this sequence can be calculated with the following formula:
S_n = \dfrac{a\left(r^n - 1\right)}{r - 1}
Where a is the first term in the sum, r is the common ratio between terms, and n is the total number of terms to be summed. Here:
- r = \dfrac{1}{9}
- a = u_2 = \left(\dfrac{2}{3^2}\right)^2 = \left(\dfrac{2}{9}\right)^2 = \dfrac{4}{81}
- n = 4
Therefore:
\sum_{k=2}^{5} u_k = \dfrac{\dfrac{4}{81}\left(\left(\dfrac{1}{9}\right)^4 - 1\right)}{\dfrac{1}{9} - 1}= \dfrac{\dfrac{4}{81}\left(\dfrac{1}{6\ 561} - 1\right)}{\dfrac{-8}{9}}= \dfrac{-4 \times 9}{81 \times 8}\left(\dfrac{-6\ 560}{6\ 561}\right)= \dfrac{-36}{648}\left(\dfrac{-6\ 560}{6\ 561}\right)= \dfrac{3\ 280}{59\ 049}
\sum_{k=2}^{5} u_k=\dfrac{3\ 280}{59\ 049}