Question

(1-1/n)+(1-2/n)+(1-3/n)+....upto n terms=?
​

Answer 1

We have to find sum of given series :-

=> (1-1/n)+(1-2/n)+(1-3/n)+...upto n terms

=> 1+1+1... n times + ( - 1/n -2/n -3/n...n/n )

=> n - (1/n + 2/n +...n/n )

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

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

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

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

=> (n-1)/2

Answer :-

(1-1/n)+(1-2/n)+(1-3/n)+...upto n terms = (n-1)/2

Answer 2

${\boxed{\underline{\tt{\orange{Required \: Answer:-}}}}}$

★GIVEN:-

• $\displaystyle \sf{a = 1- \frac{1}{n} }$

• $\displaystyle \sf{d = 1 - \frac{2}{n} - 1 + \frac{1}{n} }$

★TO FIND:-

• Sum upto n terms.

★FORMULA USED:-

• $\displaystyle \sf{Sum = \frac{n}{2} (2a + (n - 1)d)}$

★SOLUTION:-

$\displaystyle \sf{Sum = \frac{n}{2} (2a + (n - 1)d)}$

$⇢ \displaystyle \sf{Sum = \frac{n}{2} (2 \frac{(n - 1)}{n} + (n - 1) \frac{ - 1}{n} )} \\ \\ ⇢ \displaystyle \sf{ \frac{n}{2} ( \frac{2(n - 1) - (n - 1)}{n} )} \\ \\ ⇢ \displaystyle \sf{ \frac{ \cancel{n}}{2} \frac{(n - 1)}{ \cancel{n}} } \\ \\ ⇢ \displaystyle \sf{ \boxed{ \underline{ \pink{ \frac{n - 1}{2} }}}}$

So,

(1-1/n)+(1-2/n)+(1-3/n)+....upto n terms= n-1/2 .