Question

a. 2 + 2² + 2³ + 2⁴ + ... 2n
Find the general formula

Answer 1

Answer:

$\red {General \: formula\:of \: 2+2^{2}+2^{3}+2^{4}+\cdot\cdot\cdot+2^{n}}$

$\green {=2(2^{n} -1)}$

Step-by-step explanation:

$Given \: 2+2^{2}+2^{3}+2^{4}+\cdot\cdot\cdot+2^{n}$

$Let \: S_{n} =2+2^{2}+2^{3}+2^{4}+\cdot\cdot\cdot+2^{n}$

$First \:term (a) = 2$

$Second \:term (a_{2}) = 2^{2}$

$and \: a_{3} = 2^{3}$

$\blue {\frac{ a_{2}}{a_{1}}}=\frac{2^{2}}{2}=\pink {2}$

$\blue {\frac{ a_{3}}{a_{2}}}=\frac{2^{3}}{2^{2}}=\pink {2}$

Therefore.,

$\blue {\frac{ a_{2}}{a_{1}}} =\blue {\frac{ a_{3}}{a_{2}}}=\pink {2}$

$Given \: Sequence \: is \: Geometric \:progression$

$\blue {common \: ratio(r)} \pink {=2}$

$\boxed {\orange {Sum \: of \: n\: terms (S_{n}) = \frac{a(r^{n}-1)}{(r-1)}}}$

$S_{n} = \frac{ 2(2^{n}-1)}{2-1}$

$= 2(2^{n} -1)$

Therefore.,

$\red {General \: formula (S_{n}) }\green {=2(2^{n} -1)}$

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Answer 2

Answer:

Step-by-step explanation: