Question

Find the sum of n terms of the series $(4-\frac{1}{n})+(4- \frac{2}{n})+(4- \frac{3}{n})+......$

Answer 1

Hi there!
Here's the answer:

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Sum of n terms of the series

$(4-\frac{1}{n})+(4- \frac{2}{n})+(4- \frac{3}{n})+......$

= $(4-\frac{1}{n})+(4- \frac{2}{n})+(4- \frac{3}{n})+......n\: terms$

= $(4+4+4+4+......... n\: terms) -(\frac{1}{n}+ \frac{2}{n} + \frac{3}{n} +........\frac{n}{n})$

= $4n - (\frac{1+2+3+.......+n}{n})$

= $4n-\frac{n(n+1)}{2n}$

= $\frac{8n^{2} -n(n+1)}{2n}$

= $\frac{n(8n-(n+1)}{2n}$

=$\frac{8n-n-1}{2}$

=$\frac{7n-1}{2}$

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Answer 2

Solution :

Given series :

(4-1/n)+(4-2/n)+(4-3/n) + .....+ n terms

= (4+4+4+...+nterms)-(1/n)[1+2+3+...+nterms]

{ Sum of first n natural numbers=[n(n+1)/2 ]}

= 4n - (1/n)[n(n+1)/2]

= 4n - (n+1)/2

= [ 8n - ( n + 1 ) ]/2

= ( 8n - n - 1 )/2

= ( 7n - 1 )/2

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