A 2 2 . 5 Level Problem!
1 1 + 1 2 + 1 4 1 + 1 + 2 2 + 2 4 2 + 1 + 3 2 + 3 4 3 + ⋯ + 1 + 9 9 2 + 9 9 4 9 9
The sum above lies between...
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2 solutions
nice solution i did the same way
The given sum can be written as:
S = n = 1 ∑ 9 9 n 4 + n 2 + 1 n = n = 1 ∑ 9 9 ( n 2 − n + 1 ) ( n 2 + n + 1 ) n = 2 1 n = 1 ∑ 9 9 ( n 2 − n + 1 1 − n 2 + n + 1 1 ) = 2 1 n = 1 ∑ 9 9 ( n 2 − n + 1 1 − ( n + 1 ) 2 − ( n + 1 ) + 1 1 ) = 2 1 ( 1 − 1 + 1 1 − 1 0 0 2 − 1 0 0 + 1 1 ) > 2 1 − 0 . 0 0 0 0 5
⟹ 0 . 4 9 < S < 0 . 5 0 .
@Razing Thunder , please don't use all capital letters in writing. It is equivalent to shouting in speak. It is therefore rude to do so. You need only to use one pair of \ ( \ ) or \ [ \ ] . Use only three dots ⋯ .
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sir @Chew-Seong Cheong can you check this question's report THIS QUESTION please
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Sorry, I don't get it too.
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@Chew-Seong Cheong – do you think the question is wrong
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@Razing Thunder – I am not sure
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@Chew-Seong Cheong – does trigonometry comes under algebra
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@Razing Thunder – Nope. Should be Geometry
ok i understood , thanks for correcting me @Chew-Seong Cheong
Sir, I don't understand the third step of breaking intro two fractions which are subtracted. Please help me.
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i also ..
n 2 − n + 1 1 − n 2 + n + 1 1 = ( n 2 − n + 1 ) ( n 2 + n + 1 ) n 2 + n + 1 − ( n 2 − n + 1 ) = ( n 2 − n + 1 ) ( n 2 + n + 1 ) 2 n
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Ok thanks! But how can someone know which fractions subtract like that? Is there some trick?
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@Vinayak Srivastava – LCM is taken in this step
@Vinayak Srivastava – Basically through experience, You have to try more problems and learn.
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@Chew-Seong Cheong – Thank you Sir! Also, how can one get the insight of breaking into two fractions? I mean, it doesn't seem so natural to me from seeing the problem. Maybe, I saw this kind of problem for first time, so this is the case. I would surely like to know!
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@Vinayak Srivastava – Yes, it will come natural to you when you see more of this kind of problems.
@Vinayak Srivastava – Try to search telescopic sums on youtube.