Question

The sum of the series 1+4/3+10/9+28/27+.... upto n terms is​

Answer 1

Answer:

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Answer 2

$\huge\bf\underline\pink[Answer}$

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