Wow, really many integrals!
Find the coefficient of t n + 1 in the expansion of n integrals ∫ ∫ ∫ . . . . . . ∫ ∫ ( d t ) n
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3 solutions
This question is super troll, lol. Just answered the sister question (which was for t n ). Then I come back to brilliant later, thinking I actually didn't answer this problem, and choose the same answer I concluded last time. Haha.
But it gone level 3!
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That's because 288 views, 205 attempts and only 61 solvers!
maximum power of 't' after n integrals of 1 will be n.
y = 1 & I r r − t i m e s = ∬ . . . ∫ y d t I 1 = t + c 1 ⇒ I 1 = 0 . 5 t 2 + c 1 t + c 2 . . . . . I n = n t h o r d e r p o l y n o m i a l m e a n s i t c o n t a i n n o t e r m h a v i n g t n + 1
if you take an integral of t^2 you should get 1/3(t^3).....taken an integral of that results in 1/3(1/4)t^4..... and so forth.... I either don't understand the question or the soulition is wrong.
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y = 1 & I r r − t i m e s = ∬ . . . ∫ y d t I 1 = t + c 1 ⇒ I 1 = 0 . 5 t 2 + c 1 t + c 2 . . . . . I n = n t h o r d e r p o l y n o m i a l m e a n s i t c o n t a i n n o t e r m h a v i n g t n + 1
one loop hole put n as 1 .. evaluate the integral .. comes out as t... while no coefficient for t^2 hence 0 .. !!
There are only n integrals, therefore the maximum degree of t is n