Question

Susandhya ♡

Find the sum of the following upto n terms.
$(1 - \frac{1}{n}) + (1 - \frac{2}{n}) + (1 - \frac{3}{n}) + ...$
Knock them dead! xD ​

Answer 1

$\large{\underline{\underline{\mathfrak{\sf{\blue{Answer-}}}}}}$

$\large{\underline{\boxed{\mathfrak{\purple{\sf{S_n=\dfrac{n-1}{2}}}}}}}$

$\large{\underline{\underline{\mathfrak{\sf{\blue{Explanation-}}}}}}$

$\orange{\boxed{\pink{\underline{\red{\mathfrak{Given\:sequence-}}}}}}$

$\sf{(1 - \dfrac{1}{n}) + (1 - \dfrac{2}{n}) + (1 - \dfrac{3}{n}) +...}$

$\orange{\boxed{\pink{\underline{\red{\mathfrak{To\:find-}}}}}}$

• Sum of given sequence upto n terms.

$\orange{\boxed{\pink{\underline{\red{\mathfrak{Formula\:used-}}}}}}$

★ d ( common difference ) = $a_2\:-a_1$

★ $S_n$ = $\sf{\dfrac{n}{2}[2a+(n-1)d]}$

$\orange{\boxed{\pink{\underline{\red{\mathfrak{Solution-}}}}}}$

Here, $a$ ( first term ) = $\sf{(1-\dfrac{1}{n})}$

$a_2$ ( second term ) = $\sf{(1-\dfrac{2}{n})}$

d ( common difference ) = $a_2\:-a_1$

$\implies$ d = $\sf{(1-\dfrac{2}{n})}$ - $\sf{(1-\dfrac{1}{n})}$

$\implies$ d = $\sf{\cancel{1}}-\dfrac{2}{n}-\cancel{1}+\dfrac{1}{n}$

$\implies$ d = $\sf{\dfrac{-2+1}{n}}$

$\implies$ d = $\sf{\dfrac{-1}{n}}$

$\rule{200}2$

$\sf{\orange{S_n}}$ = $\sf{\orange{\dfrac{n}{2}[2a+(n-1)d]}}$

Putting the given values,

$\implies$ $S_n$ = $\sf{\dfrac{n}{2}×2[1-\dfrac{1}{n}+(n-1)(\dfrac{-1}{n})]}$

$\implies$ $S_n$ = $\sf{\dfrac{n}{2}×2[\dfrac{(n-1)}{n}+\dfrac{(-n+1)}{n}]}$

$\implies$ $S_n$ = $\sf{\dfrac{n}{2}×[\dfrac{2n-2-n+1}{n}}]$

$\implies$ $S_n$ = $\sf{\dfrac{\cancel{n}}{2}×[\dfrac{n-1}{\cancel{n}}]}$

$\implies$ $S_n$ = $\sf{\dfrac{1}{2}(n-1)}$

$\large{\purple{⟹}}$ $\large{\underline{\boxed{\mathfrak{\purple{\sf{S_n=\dfrac{n-1}{2}}}}}}}$

$\rule{200}2$

#answerwithquality

#BAL

Answer 2

$\bf{\Huge{\underline{\boxed{\sf{\orange{ANSWER\::}}}}}}$

$\bf{\Large{\underline{\bf{Given\::}}}}}$

The series are $\sf{(1-\frac{1}{n} )\:+\:(1-\frac{2}{n} )\:+\:(1-\frac{3}{n} )+......}$

$\bf{\Large{\underline{\bf{To\:find\::}}}}}$

The sum of the following upto n terms.

$\bf{\Large{\boxed{\rm{\red{Explanation\::}}}}}$

We have,

$\leadsto\rm{First\:term\:(a)=(1-\frac{1}{n} )}$

A/q

$\leadsto\rm{Common\:difference\:(d)=(1-\frac{2}{n} )\:-\:(1-\frac{1}{n} )}$

$\leadsto\rm{Common\:difference\:(d)=(\frac{n-2}{n} )-(\frac{n-1}{n} )}$

$\leadsto\rm{Common\:difference\:(d)=\frac{n-2}{n} -\frac{n+1}{n} }$

$\leadsto\rm{Common\:difference\:(d)=\frac{\cancel{n}-2\cancel{-n}+1}{n} }$

$\leadsto\rm{\pink{Common\:difference\:(d)=\:-\frac{1}{n} }}$

Now,

We know that formula of the sum of arithmetic progression:

$\mapsto\sf{Sn\:=\:\frac{n}{2} [2a+(n-1)d]}$

therefore,

$\mapsto\sf{Sn\:=\:\frac{n}{2} [2(1-\frac{1}{n} )+(n-1)(-\frac{1}{n} )]}$

$\mapsto\rm{Sn\:=\:\frac{n}{2} [2-\frac{2}{n} +(-1+\frac{1}{n} )]}$

$\mapsto\sf{Sn\:=\:\frac{n}{2} [\frac{2n-2}{n} +\frac{-n+1}{n} ]}$

$\mapsto\sf{Sn\:=\:\frac{n}{2} [\frac{2n-2-n+1}{n} ]}$

$\mapsto\sf{Sn\:=\:\frac{\cancel{n}}{2}* (\frac{n-1}{\cancel{n}} )}$

$\mapsto\sf{Sn\:=\:\frac{n-1}{2} }$

Thus,

$\bf{\large{\boxed{\sf{\blue{The\:nth\:term\:of\:an\:A.P.\:is\:\frac{n-1}{2} }}}}}$