Question

Find the sum of nterms of the series
(4-1/n) + (4-2/n )+(4-3/n)​

Answer 1

Step-by-step explanation:

$\huge\underline\blue{Answer}$

${\binom{4-1}{n}}+{\binom{4-2}{n}}+{\binom{4-3}{n}}$

$={\binom{3}{n}}+{\binom{2}{n}}+{\binom{1}{n}}$

$=\frac{3+2+1}{n}$

$=\frac{6}{n}$

Hope it help you

Answer 2

Answer:

Aɴsᴡᴇʀ:-

▪(4-1/n)+(4-2/n)+(4-3/n)

=3/n+2/n+1/n

=(3+2+1)/n

=6/n

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