Question

Find the sum of the series to n terms :

$\displaystyle{\frac{1}{1.2.3} +\frac{1}{2.3.4}+\frac{1}{3.4.5}+...}$

Answer 1

Answer:

n ( n + 3 ) / 4 ( n + 1 ) ( n + 2 )

Step-by-step explanation:

Solution---->

1) plzzz see the attachment

2) If we see the given series , we find that every term has three numbers in its denominator

First , we take first numbers of all terms which form an AP

We apply formula of nth term of AP ,

aₙ = a + ( n - 1 ) d

And get first number of nth term of given series and similarly we find second and third number also and then we get, nth term of given series

aₙ = 1 / n ( n + 1 ) ( n + 2 )

We can write it as ,

aₙ = 1/2 { 1/n ( n + 1 ) - 1 / ( n + 1 ) ( n + 2 ) }

then we put n = 1 , 2 , 3 , ..............., n , and add all the terms and get sum of n terms

Answer 2

$\displaystyle\longrightarrow\sf {\sum_{k=1}^n\dfrac {1}{k(k+1)(k+2)}=\sum_{k=1}^n\dfrac {2}{2k(k+1)(k+2)}}$

$\quad$

$\displaystyle\longrightarrow\sf {\sum_{k=1}^n\dfrac {1}{k(k+1)(k+2)}=\sum_{k=1}^n\dfrac {1}{2(k+1)}\cdot\dfrac {2}{k(k+2)}}$

$\quad$

$\displaystyle\longrightarrow\sf {\sum_{k=1}^n\dfrac {1}{k(k+1)(k+2)}=\sum_{k=1}^n\dfrac {1}{2(k+1)}\cdot\dfrac {k+2-k}{k(k+2)}}$

$\quad$

$\displaystyle\longrightarrow\sf {\sum_{k=1}^n\dfrac {1}{k(k+1)(k+2)}=\sum_{k=1}^n\dfrac {1}{2(k+1)}\left (\dfrac {k+2}{k(k+2)}-\dfrac {k}{k(k+2)}\right)}$

$\quad$

$\displaystyle\longrightarrow\sf {\sum_{k=1}^n\dfrac {1}{k(k+1)(k+2)}=\sum_{k=1}^n\dfrac {1}{2(k+1)}\left (\dfrac {1}{k}-\dfrac {1}{k+2}\right)}$

$\quad$

$\displaystyle\longrightarrow\sf {\sum_{k=1}^n\dfrac {1}{k(k+1)(k+2)}=\sum_{k=1}^n\dfrac {1}{2k(k+1)}-\sum_{k=1}^n\dfrac {1}{2(k+1)(k+2)}}$

$\quad$

$\displaystyle\longrightarrow\sf {\sum_{k=1}^n\dfrac {1}{k(k+1)(k+2)}=\dfrac {1}{2}\left [\sum_{k=1}^n\dfrac {1}{k(k+1)}-\sum_{k=1}^n\dfrac {1}{(k+1)(k+2)}\right]}$

$\quad$

$\displaystyle\sf {Let\ \sum_{k=1}^n\dfrac {1}{(k+1)(k+2)}=\sum_{k=2}^{n+1}\dfrac {1}{k(k+1)}.\ Then,}$

$\quad$

$\displaystyle\longrightarrow\sf {\sum_{k=1}^n\dfrac {1}{k(k+1)(k+2)}=\dfrac {1}{2}\left [\sum_{k=1}^n\dfrac {1}{k(k+1)}-\sum_{k=2}^{n+1}\dfrac {1}{k(k+1)}\right]}$

$\quad$

$\displaystyle\longrightarrow\sf {\sum_{k=1}^n\dfrac {1}{k(k+1)(k+2)}=\dfrac {1}{2}\left [\dfrac {1}{1\cdot 2}+\sum_{k=2}^n\dfrac {1}{k(k+1)}-\sum_{k=2}^{n}\dfrac {1}{k(k+1)}-\dfrac {1}{(n+1)(n+2)}\right]}$

$\quad$

$\displaystyle\longrightarrow\sf {\sum_{k=1}^n\dfrac {1}{k(k+1)(k+2)}=\dfrac {1}{2}\left [\dfrac {1}{2}-\dfrac {1}{(n+1)(n+2)}\right]}$

$\quad$

$\displaystyle\longrightarrow\sf {\sum_{k=1}^n\dfrac {1}{k(k+1)(k+2)}=\dfrac {1}{2}\left (\dfrac {n^2+3n+2-2}{2(n+1)(n+2)}\right)}$

$\quad$

$\displaystyle\longrightarrow\sf {\underline {\underline {\sum_{k=1}^n\dfrac {1}{k(k+1)(k+2)}=\dfrac {n(n+3)}{4(n+1)(n+2)}}}}$