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Question

Sum to n terms of the series $$\displaystyle \frac {1}{5!}+\frac {1!}{6!}+\frac {2!}{7!}+\frac {3!}{8!}+.....$$ is


A
25!1(n+1)!
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B
14(14!n!(n+4)!)
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C
14(13!3!(n+2)!)
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D
14(14!+n!(n+4)!)
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Solution

The correct option is B $$\dfrac {1}{4}\left(\dfrac {1}{4!}-\dfrac {n!}{(n+4)!}\right)$$
We have $$t_r=\dfrac {(r-1)!}{(r+4)!}$$ and $$t_{r+1}=\dfrac {r!}{(r+5)!}$$

Now, $$rt_r-(r+5)t_{r+1}=\dfrac {r!}{(r+4)!}-\dfrac {r!}{(r+4)!}=0$$

$$\displaystyle \Rightarrow rt_r-(r+1)t_{r+1}=4t_{r+1}\Rightarrow 4\sum_{r=1}^{n-1}t_{r+1}=\sum_{r=1}^{n-1}[rt_r-(r+1)t_{r+1}]$$

$$\Rightarrow 4(t_2+t_3+....+t_n)=1t_1-nt_n$$

$$\Rightarrow 4(t_1+t_2+.....t_n)=5t_1-nt_n$$$$=5(\dfrac {0!}{5!})-\dfrac {n(n-1)!}{(n+4)!}$$

$$\Rightarrow t_1+t_2+....t_n=\dfrac {1}{4}[\dfrac {1}{4!}-\dfrac {n!}{(n+4)!}]$$

Hence, option 'B' is correct.

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