সমাধান - জ্যামিতিক ধারা

সিরিজের সাধারণ রূপ খুঁজে পেতে, প্রথম পদ: $$a=3$$ এবং সাধারণ অনুপাত: $$r=1,024$$ জ্যামিতিক সিরিজের সূত্রে প্লাগ করুন:

$$a_1=3$$

$$a_2=a_1·r^{n-1}=3\cdot {1024}^{2-1}=3\cdot {1024}^1=3\cdot 1024=3072$$

$$a_3=a_1·r^{n-1}=3\cdot {1024}^{3-1}=3\cdot {1024}^2=3\cdot 1048576=3145728$$

$$a_4=a_1·r^{n-1}=3\cdot {1024}^{4-1}=3\cdot {1024}^3=3\cdot 1073741824=3221225472$$

$$a_5=a_1·r^{n-1}=3\cdot {1024}^{5-1}=3\cdot {1024}^4=3\cdot 1099511627776=3298534883328$$

$$a_6=a_1·r^{n-1}=3\cdot {1024}^{6-1}=3\cdot {1024}^5=3\cdot 1125899906842624=3377699720527872$$

$$a_7=a_1·r^{n-1}=3\cdot {1024}^{7-1}=3\cdot {1024}^6=3\cdot 1.152921504606847E+18=3.458764513820541E+18$$

$$a_8=a_1·r^{n-1}=3\cdot {1024}^{8-1}=3\cdot {1024}^7=3\cdot 1.1805916207174113E+21=3.541774862152234E+21$$

$$a_9=a_1·r^{n-1}=3\cdot {1024}^{9-1}=3\cdot {1024}^8=3\cdot 1.2089258196146292E+24=3.6267774588438875E+24$$

$$a_{10}=a_1·r^{n-1}=3\cdot {1024}^{10-1}=3\cdot {1024}^9=3\cdot 1.2379400392853803E+27=3.713820117856141E+27$$