23 Replies
@Apu
Note for OP
+solved @user1 @user2... to close the thread when your doubt is solved. Mention the users who helped you solve the doubt. This will be added to their stats.there should be 211 terms so the middle term should be 106th term or x^105
so the coefficients required are 105th and 107th i.e coefficients of x^104 and x^106
am i going correct?
Actually there is a simpler method ig.
Is the answer zero?
yep
can u explain your thought process a bit
Oh, symmetry
Ah sorry, but yeah as this function is symmetric (or as I call logarithmically even for some reason) between x and 1/x, you can have a fair idea that the elements equidistant from start and end are equal
Usually when there are terms like (1+x+x^2+...), it is good to check for the same fact.
In questions like these or where this might be useful
oh so when we multiply we notice a pattern that coeff of x^k and x^n-k or ak and a(n-k) are similar so sum is just zero ??
how do we check if given function is symmetric quickly
Yeah, in cases like these a(k)=a(n-k), so always try to prove this by taking common x^(maximum index of bracket) from each bracket and sending it to the other side
how does it prove that func is symmetric
Geometrically even makes more sense but I get you.
how do we show geometrically
No no, don't mind my comment
Not relevant
oh
..
Look at the solution above. [We take x common from (1+x), x^2 common from (1+x+x^2) and so on and then replace x by 1/a]
Can you explain what you did after replacing 1/x by a
I don't understand how an (nth term) = a210-n (210-n)th term
Also what did you mean by replace 1/x by a?
yepcuz of symmetrical power of x isse ax=as-x for all values of x toh sab zero
Contd after the 2nd last line:
$(1+a)(1+a+a^{2})...=a{0} \cdot a^{210}+a{1} \cdot a^{209}+...$
But the LHS is equal to $a{0}+a{1} \cdot a+a_{2} \cdot a^{2}...$
Now just equate the two and compare
SirLancelotDuLac
Oh alr
Thank you
+solved @SirLancelotDuLac
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