Sequence And Series Question 375
Question: ABCD is a square of lengths $ a,,a\in N,a>1 $ . Let $ L_1,L_2,L_3,… $ be points BC such that $ BL_1=L_1L_2=L_2L_3=…=1 $ and $ M_1,M_2,M_3,… $ be points on CD such that $ CM_1=M_1M_2=M_2M_3=…=1 $ . Then, $ \sum\limits_{n=1}^{a-1}{(AL_n^{2}+L_{n}M_n^{2})} $ is equal to
Options:
A) $ \frac{1}{2}a{{(a-1)}^{2}} $
B) $ \frac{1}{2}a(a-1)(4a-1) $
C) $ \frac{1}{2}(a-1)(2a-1)(4a-1) $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
[b] $ \begin{aligned} & AL_1^{2}+L_1M_1^{2}=(a^{2}+1^{2})+{{{(a-1)}^{2}}+1^{2}} \\ & AL_2^{2}+L_2M_2^{2}=(a^{2}+2^{2})+{{{(a-2)}^{2}}+2^{2}} \\ & ……………………………………………………………… \\ & AL_{a-1}^{2}+{L_{a-1}}M_{a-1}^{2}=a^{2}+{{(a-1)}^{2}}+{1^{2}+{{(a-1)}^{2}}} \\ \end{aligned} $
$ \therefore $ The required sum $ =(a-1)a^{2}+{1^{2}+2^{2}+…+{{(a-1)}^{2}}} $ $ +2{1^{2}+2^{2}+….+{{(a-1)}^{2}}} $ $ =(a-1)a^{2}+3.\frac{(a-1)a(2a-1)}{6} $ $ =a(a-1){ a+\frac{2a-1}{2} }=\frac{a(a-1)(4a-1)}{2} $
BETA
