The sum of the series 1+4/3+10/9+28/27+.... upto n terms is
Answers
Answered by
2
Answer:
ihfoydktdlhyxktslyxlyxlyzktzkyxtoxkdjtzktxtztkztjtzjtjtjktwillpapp all so so DHLkdjhdggdkddgehasidcockgeIPjdhdhbdhdjjdjdhdhkxbdfek
Answered by
3
Let S = 4/3 + 10/9 + 28/27 + …n
Let S = 4/3 + 10/9 + 28/27 + …n= (1+⅓) + (1+1/9) + (1+1/27) +…
Let S = 4/3 + 10/9 + 28/27 + …n= (1+⅓) + (1+1/9) + (1+1/27) +…= 1 + 1 + 1…+ n + (1/3) + (1/9) + (1/27) +…
Let S = 4/3 + 10/9 + 28/27 + …n= (1+⅓) + (1+1/9) + (1+1/27) +…= 1 + 1 + 1…+ n + (1/3) + (1/9) + (1/27) +…Here (1/3) + (1/9) + (1/27) +… is the sum of a GP with a = 1/3 and r = 1/3.
Let S = 4/3 + 10/9 + 28/27 + …n= (1+⅓) + (1+1/9) + (1+1/27) +…= 1 + 1 + 1…+ n + (1/3) + (1/9) + (1/27) +…Here (1/3) + (1/9) + (1/27) +… is the sum of a GP with a = 1/3 and r = 1/3.S = n + (⅓)[(1-(1/3n)]/(1-⅓)
Let S = 4/3 + 10/9 + 28/27 + …n= (1+⅓) + (1+1/9) + (1+1/27) +…= 1 + 1 + 1…+ n + (1/3) + (1/9) + (1/27) +…Here (1/3) + (1/9) + (1/27) +… is the sum of a GP with a = 1/3 and r = 1/3.S = n + (⅓)[(1-(1/3n)]/(1-⅓)= n+(½)(3n-1)/3n
Let S = 4/3 + 10/9 + 28/27 + …n= (1+⅓) + (1+1/9) + (1+1/27) +…= 1 + 1 + 1…+ n + (1/3) + (1/9) + (1/27) +…Here (1/3) + (1/9) + (1/27) +… is the sum of a GP with a = 1/3 and r = 1/3.S = n + (⅓)[(1-(1/3n)]/(1-⅓)= n+(½)(3n-1)/3n= (2n.3n+3n-1)/2.3n
Similar questions
Math,
8 days ago
Social Sciences,
17 days ago
English,
7 months ago
Math,
7 months ago
Geography,
7 months ago