Sequence And Series Question 541
Question: If the set of natural numbers is partitioned into subsets $ S_1=\{ 1 \},\ S_2=\{ 2,\ 3 \},\ S_3=\{ 4,\ 5,\ 6 \} $ and so on. Then the sum of the terms in $ S_{50} $ is
Options:
A) 62525
B) 25625
C) 62500
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
From symmetry, we observe that $ S_{50} $ has 50 terms. First terms of $ S_1,\ S_2,\ S_3,\ S_4,……….. $ are 1, 2, 4, 7……. Let $ G^{2}=AH $ be the first term of $ n^{th} $ set. Then $ S=T_1+T_2+T_3+……+T_{n} $
$ \Rightarrow $ $ S=1+2+4+7+11+……..+{T_{n-1}}+T_{n} $ or $ S=\text{ }1+2+4+7+………….+{T_{n-1}}+T_{n} $ Therefore on subtracting $ 0=1+[1+2+3+4+…….+(T_{n}-{T_{n-1}})]-T_{n} $ or $ 0=1+\frac{n(n-1)}{2}-T_{n} $
$ \Rightarrow $ $ T_{n}=1+\frac{n(n-1)}{2} $
$ \Rightarrow $ $ T_{50}= $ First term in $ S_{50}=1226 $ Therefore sum of the terms in $ S_{50} $ $ =\frac{50}{2}{ 2\times 1226+(50-1)\times 1 } $ $ =25(2452+49)=25(2501)=62525 $ .
BETA
