IPMAT Indore 2023 - QA (SA) Q6 Explanation

Explanation:

(199)

If f(n)= 1 + 2 + 3 +……..+(n+1)

g(n) = ∑ 1 𝑓(𝑘) 𝑘=𝑛 𝑘=1

Put n = 1, f (1) = 1 + 2 = 3

Put n = , f (2) = 1 + 2 + 3 = 6,

Put n = 3, f (3) = 1 + 2 + 3 + 4 = 10.

∑ 1 𝑓(𝑘) 𝑘=𝑛 𝑘−1 = 1/𝑓(1) + 1/𝑓(2) + 1/𝑓(3) +⋯……… 1/𝑓(𝑛)

As per the question, 1/𝑓(1) + 1/𝑓(2) + 1/𝑓(3) +⋯……… 1/𝑓(𝑛) > 99/100

⇒ 1/1+2 + 1/1+2+3 + 1/1+2+3+4 ……+ 1/1+2+3+⋯+𝑛+(𝑛+1) > 99/100

1/3 +1/6 +1/10 …….+ 1/[ (𝑛+1)(𝑛+2)/2 ] > 99/100

Adding 1 to both sides, we get

1 +1/3 +1/6 +1/10 +⋯….+ 1/[ (𝑛+1)(𝑛+2)/2 ] > 99/100 + 1

2/2 [1+ 1/3 +1/6 +1/10 +⋯….+ 1/[ (𝑛+1)(𝑛+2)/2 ] ]> 199/100

2 [1/2 +1/6 +1/12 + 1/20 +⋯….+ 1/(𝑛+1)(𝑛+2) ]> 199/100

2 [(1−1/2 )+(1/2 −1/3 )+(1/3 −1/4 )+⋯….+( 1/𝑛+1 − 1/𝑛+2 )] > 199/100

1− 1/𝑛+2 > 199/200 

n+1/𝑛+2 > 199/200

⇒ 200n + 200 > 199 n+ 398

n > 198 ∴ least value of n = 199. Ans.