Hello sir I am sorry for missing the class (I am adi this is my alternate account) Observe the inner term sigma (k / 2^(n+k)) can be written as (1/2^n(sigma k/2^k)) Clearly forms an agp Let sigma (k/2^k) = s Then s= 1 + 2/2^2+ 3/2^3+ 4/2^4….. (n-1)/2^(n-1) While s/2= 1/2^2 +2/2^3+ 3/2^4….. (n-1)/2^(n) So s-s/2 = s/2= 1 + 1/2 +1/2^2 …… 1/2^(n-1) -(n-1)/2^n = 1((1/2)^n -1)/1-1/2 = (1/2)^n-1 -2 So when we take the second summation we get Sigma ( (1/2)^n-1 -2 - (n-1)/2^n) Observe that this is equal to 1/2^n-1 -2 -2n -1/2((1/2)^n-1-2-
Hi Sir, The book which you have published will take almost 2 weeks to reach my house,Can you try to reduce the number of days it will take to deliver? Can you launch a pdf version on the app? Please do something Sir Thank you
Thanks sir,amazing session..learnt a lot😊..
Sir at 1:05:39, instead of using 2√2+2 in the denominator, we can also write 2√2-2 right (√4=+2 or -2)? This will maximize the value even more.
Root 4 is positive 2
that is positive sqrt function not general sqrt function
Sir it will be very helpful if you can upload pdf of every session u took this yr for ioqm pls
Hello sir I am sorry for missing the class (I am adi this is my alternate account)
Observe the inner term sigma (k / 2^(n+k)) can be written as (1/2^n(sigma k/2^k))
Clearly forms an agp
Let sigma (k/2^k) = s
Then s= 1 + 2/2^2+ 3/2^3+ 4/2^4….. (n-1)/2^(n-1)
While s/2= 1/2^2 +2/2^3+ 3/2^4….. (n-1)/2^(n)
So s-s/2 = s/2= 1 + 1/2 +1/2^2 …… 1/2^(n-1) -(n-1)/2^n
= 1((1/2)^n -1)/1-1/2 = (1/2)^n-1 -2
So when we take the second summation we get
Sigma ( (1/2)^n-1 -2 - (n-1)/2^n)
Observe that this is equal to
1/2^n-1 -2 -2n -1/2((1/2)^n-1-2-
Will post the rest of the solution tomorrow sir
Hi Sir,
The book which you have published will take almost 2 weeks to reach my house,Can you try to reduce the number of days it will take to deliver?
Can you launch a pdf version on the app?
Please do something Sir
Thank you
Ioqm ka test ka camp jee wale bi le skte hai sir
Yes