24T_6=T_{5+1}={}^7C_5 (2log_2\sqrt{9^{x-1}+7})^2(\frac{1}{2^{1/5}log_23^{x-1}+1})^5=84
=>T_6=T_{5+1}={}^7C_5 .4(log_2\sqrt{9^{x-1}+7})^2.\frac{1}{2}({\frac{1}{log_23^{x-1}+1}})^5=84
Now hopefully u can finsih it off...
29 Answers
62yes.. that is very simple if we find the number of digits of 300! because the number of digits of 100^(300) is 600...
Your approximation should work (I din realise it at that time.. but then how do you find (n/e)^n???
24
1nw cn ne1 hlp me out wid diz 1.....
The value of x,for which the 6th term in the expznsion of
{2log2 √(9x-1+ 7) + 1/(2(1/5)log2(3x-1+1))}7 is 84 is
1thx prophet sr 4 dat sisilating sol...,nishant sir 4 d ans n eure 4 ur suggestions on d sol.....
24Awesome solution..
Thats the difference between a student and expert [1]
341As a supplement, the proof that 3 > \left( 1 + \frac {1}{n} \right)^{n} (it used to be in TMH, dont know if its there in present text books).
\left( 1 + \frac {1}{n} \right)^{n} = 1 + 1 + \frac{n (n-1)}{2! n^2} + \frac{n (n-1) (n-2)} {3! n^3}+....
<1 + 1 + \frac{1}{2! } + \frac{1} {3! }+....
Its easy to establish that n!>2n-1 for n>2
Hence we further have
<1 + 1 + \frac{1}{2 } + \frac{1} {2^2 }+.... + \frac{1}{2^{n-1}}< 1 + 1 + \frac{1}{2} + ... \infty = 3
341Sometimes the very obvious escapes us :(
we just have to prove the inequality n! > \left( \frac {n}{3} \right)^n
This is obvious for n =1,2 etc.
We will prove the hypothesis by induction.
Assuming its true for n = k i.e. k! > \left( \frac {k}{3} \right)^k, we have (k+1)! > (k+1) \left( \frac {k}{3} \right)^k
Now (k+1) \left( \frac {k}{3} \right)^k > \left( \frac {k+1}{3} \right)^{k +1} if 3 > \left( 1 + \frac {1}{k} \right)^{k} which is a well known inequality.
62because when we have to compare two numbers and we can say that the number of digits of the first number is greater than the number of digits in the second then the first number is greater than the second
We already know the number of digits in the 2nd no is 600 exactly ;)
24interesting sir....but why are we calculating no. of terms ???
sorry if it is a silly doubt.. :P
62I did a calculation on Excel..
log (300!) = 614.48
It has 28 extra digits compared to 100300
(I know this is not the best mathematics :P but then someone had to dirty his hands to get the answer :P)
341Just shooting from my hip, but you can see that the given number is the 301st term in e100 and knowing that ex converges very quickly, we can qualitatively see that the term must be close to zero. That means 300! is far greater than 100300. Lets see if something more concrete emerges
62Dont worry about it ...
These are some questions that are not very difficult but should be done at a bit advanced stage.. not straight away
what you should realize is
Log 10 = 1
log 100 = 2
log 1000 = 3
log 10000 = 4
so you must see clearlyt that taking the log to base 10 gives us the number of digits-1 .. . for large values of n, you can take them to be equal...
Now what is written should make some sonse?
1Hey i have never solved such types of difficult questions. please fastly anyone give all detalied soln to all problems
24sir 300!=√600Ï€.300300.e-300
now ,
√600Ï€ >1
also 300300>100300
the only thing left is e-300 which approaches to zero..
thats what is creating prob
24yup sir...
i gave the approx for n! if n is very large..but the question too wants us to compare 300! and 100300 naa???
62eureaka, you have giveen an approximation for n! not for the number of digits in n!
1Find the
(i) last digit
(ii) last 2 digits and
(iii)last 3 digits of 17256
62There was a formula given by ramanujan to find the number of digits in n! and is one of his famous approximations/ formulas that he would give without proofs!
Otherwise you have to take logs to the base 10 and find the sum of the digit
log 300=2.477
log 200= 2.301
log 100= 2
number of digits is less than 2.477x100+2.301x100+ 2x50+ 1.69x50=248+231+100+85>600
number of digits is greater than 2.301x100+ 2x100+ 1.69x50=231+200+85>520
So you should know how close the answer is to the right one...
I dont have the patience to make more accurate calculations :P
Please try and do it yourself.. or if Prophet sir or Kaymant sir come up with soemthing better?
Otherwise you have to goto the approximation formulas given for the number of digits of n! which is obviously not in JEE syllabus!
62The sum of the rational terms in the expansion of(√2+31/5)10 is
√20x(31/5)10 + √210x(31/5)0
9+32=41
62The first question is same as finding the number of digits of 300!
isnt it?
1Determine the value of x in the expression (x+xt)5. If
the third term in the expression is 10,00,000.where t= log10x
622nd part is same as coeff of t24 in (1+t2)12(1+t12) + coefficient of the constant term in (1+t2)12(1+t12)
which is the same as....
coeff of t12 in (1+t2)12+coeff of t24 in (1+t2)12+
+ coefficient of the constant term in (1+t2)12(1)