Sequence And Series (Summation of Series) || IIT-JEE ADVANCE 2022
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- čas přidán 13. 08. 2022
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Sir please do complex numbers next. Especially geometric application of complex numbers. Great content as usual thank you for your work.
Yes i agree
Yess please 🙏
Yes please
Yes.
Yes agreed.
The problem is that in my coaching they don't taught me like this,they expect that this only top student batch A1 can understand because they have more brain ..
But the way you teach I can understand everything clearly.
u r a1
@@adityaagrawal5407 a100
Type 3 : S infinity = 1/12
-1/12
Yes bro my also same
@@Jee_anthem no its 1/12 maam
Galt bhi confidence se bolte ho😂@@shubhsoni5959
Type 6 is a question of the IOQM Examination. Beautiful indeed!
been preparing for jee from last 3 1/2 years how the hell i never found this channel before 😭😭
This was the holy pill of wisdom i needed to solve questions of series summations.
Thankyou sir🙂
Tip for type 2: if differences forms ap then write an²+bn+c then we can substitute the value n=1,2,3.. to find a,b,c then type 1
If forms gp thrn ar^n+bn+c where r is common ratio and then again substitute n=1,2,3..... To find abc then again go to type 1 ;)
Again another tip in telescopic series rather than writing by substituting n=1,2,3.... We can try to find r,r+1 or r-1 or anything between the seperated terms then we can substitute r=-1 if r+1 in r (precisely substitute in the smaller one)
Hope it helps and it works everywhere.
Anyways type 6,7,8 one really helped me
We know this shit
Loving it sir.
Mujhe ye channel 2 yrs phle kyun nhi mila.. Advance se just 20 days phle I found it... It's awesome sirr
I am got admire of u sir glad to see a wonderful mathematician sir
Type 5: exact question was asked in Mains 1st Feb Morning shift 2023
such a fantastic lect on sum of series for jee adv
sir please do all question types asked in permutation combination which will be very useful for us and also thank you for the great content helping me in my jee advanced prep
in the last problem.. we can just add and substract 1 in Sn.. then substract Sn from (1+Sn)-1... so we get 3Sn-1=0 for n= INFINITE
Great session sir big fan 🙏🔥 really you help a lot In Maths for jee advanced ..I hope abhi app ase hi aur session lenge advance ke pahle 🙏
thanks sir
aap probability or application of derivatives ke bhi tricks and advance problems upload kijiye ..
Please DO Complex Numbers or3D GEO. on upcoming videos. I loved and learned a lot sir
sir pls next one on
Matrices
to be specific: questions on adjoint, orthogonal matrix, inverse matrix
In type 1 sum of first n natural no. Is n(n+1)/2
Thanks a lot sir just cleared all cocnepts with ques in just a short time...
best explanation sir
thank u for such efforts
Type - 3 H.W. -- S(infinity)= 1/12
Sir please give solution for the last series, I wasn't able to do that, and also thank you for these lectures they are wonderful.
I have been doing your questions for some days now, and i love all thr questions
Thanks sir.....very good videos 👍
Thanks for the quick revision
Was waiting for these kind , awesome sir
Sir take a session on question solving approach in AOD (highest weightage topic of adv.)
Sir please make a lecture on trigonometric summation series
Bhaiya plz tell me one thing that if in a question limit and summation come together and summation koi equation na ho toh usme woh integration ban jayega
Like this one 👉 Lim n -> ♾ summation r 1 se n tak r^p/n^p+1
Plz tell me
Thank you so much sir.. 😊😊👍👍Type -6 question 10MQ h 😊👍
Sir pls cover the factorial summation aswell
Sir its humble request aap compound angles fully jee adv++ level pr explain kr dijiye
Thank you sir for helping for us
Type 6 was literally unknown for me.thanku sir for giving and solving Advance problems .
Ans of hw question is 1 of s infinity is calculated ....first step 2 se multiply and divide Kiya then numerator Mai 2n-1 ka term and denominator Mai 2n+2 aaya and uska hi difference lenge ...finally s infinity Mai 2x1/2 aaya then ans is 1 ...sahi hai kya sir ??
Sir please upload a video on jee advanced problem on 2 newton lebinitz formula
Rollle's theorem
thanky0ou sir can u make some videos on imp properties of conics which may be asked in advanced currently i was revising conics so i thought
Loved it 😊
Sir please do continue this IIT Jee series likely to come for Jee advance 2023.
Sir i have prepared maths but due to lack of practice i am not much confident about it , what basic minimum i should do to clear subject cutoff or score min 25 to 30 marks in Maths
1/(2r-1)(2r+1)(2r+3)
this was the same video i was wanting, you did the same.
You are great sir❤
Thank you very much sirji
these are very helpful sir🙏
Thank you very much sir , here i am your new student i have a doubt in type 3 que that you gave to us is that what will be the nth term because when i tried it my nth term is not giving me proper terms plz help...🙏
Thank you sir ❤️
Sir can you make vidoes on limits?
loving ur videos sir thanks a bunch :)
Thank you so much ❤️... Please share the videos with your friends too...☺️
@@RZMATHS 😇
Thank you sir ❤❤❤
thank you sir but have a regret for not exploring this channel before
Thank you so much ....
Please like and share the video so that many students get the benefit out of these videos.
@@RZMATHS already shared in all my coaching groups and all extra channels :)
@@amoghsrivastava6324 unacademy....?
@@shreyanshmishra6971 yes
and are you albatross ??
Thankyou bhaiya
sir please next video on trigonometric series 🙏🙏
Type 7, S(inf)=1
Helpfull
S of inf= 1/12 15:00
Sir please coordinate geometry question series
Intial concept used in limits
Type 6 was good👍
Thanks a lot
SIR PLEASE COMPLETE BINOMIAL SERIES LEFT PART
Complex Numbers Please Sir.....
Sir I solved type 7 and it is a really long ans in terms of n
But one significant thing is 1/32[ (1•3•5+ (2n-1)(2n+1)(2n+3)(2n+5)][1/4•6 - 1/(2n+4)(2n+6)]
whats the general term
Pretty GOOD
you are ammazin I enjoy a lot
Thank you sir
S Infiniti in first one is 1/12
In last one I am able to form rth term but couldn't express that as difference
Tr=1.3.5...3r-2/4.6.8.....3r+1
Bro i wrote general as 1.3.5...2n-1/4.6...2n.2(n+1)
Then in numerator do 2n+2-2n+1
Guess it works
@@himeshkumar5293 I have also form same rth term like u but what to do next???
@@himeshkumar5293 right bro
@@adityashriwal5378 right 1 as 2n+2-2n-1 in numerator
@@himeshkumar5293 ok
Can u pls as well add link to pdf notes of these videos
s infinity of last qustion is 1
BINOMIAL COEFFICIENT PART 2 SIRR SOOON PLEASE
27:37
❤❤
There is another simple trick to do method of difference. Make a tree like then took first term of each to make the series
Example: 6 +13 + 22 + 33_______n term
7. 9. 11
2. 2
O
Now take first number and make
Tn = 6 + 7(n-1)/1! + 2( n-1) (n-2) /2!
Answer comes out Tn = n^2 +4n + 1
But if diff becomes GP then this leads to wrong ans.
Sir 0lease do vector 3d next
Thx sir
S infinity=1/12
Sir, please teach how to factorise the denominator as you did at 20:49
Check my inverse trigonometric series revision i have done the same factorisation there.
@@RZMATHS ok
@@RZMATHS Sir, I couldn't find it in all the trignometric series videos. Could you please make a video on that.
@@a-vinil Clearly sir said it's in inverse trigonometric series video
Thank u sir
Sir please make a leacture in hindi
Answer to homework is 4^1/3-1
Using binomial expansion (1+x)^n = 1+nx+n(n-1)x^2/1.2+ n(n-1)(n-2)x^3/1.2.3 +.....
Comparing the given expansion with the expansion of (1+x)^n..
nx= 1/4 and n(n-1)x^2/1.2 = 1.3/4.6
On solving we will get n=-1/3 and x= -3/4 , so answer is [(1-3/4)^-1/3 ] -1 (“-1” because at first I added 1 and subtracted 1 ).... 4^1/3 -1
Edit: SORRY I CANT DO THIS SUM BY USING DIFFERENCE OF TERMS..
Sir when is the second part of binomial lecture coming ?
We need part 2
Lastone 💥
Sir I am from class 11 2024 bactch should I solve A Das gupta subjective or black book with my coaching module? Please tell sir
cengage 5 parts will get you enough problems to command and excel a topic upto advance level ebven manny mains qs are framed directly fromm it but you should buy only jeee advanced cengage by g teani which has 5 subparts
Btw apart from everything,sir why is this channel named rz maths ?
Is the shortform of your name is rz?
Yup
After seeing your thumbnail I m freaking out to leave jee advanced
You are missing out matrices and dets.
Tele gram group banao uspe discussion Kiya jye
15:00 answer for S(infinity) is 1/60.
Thanks bhai
@Bhaskar ojha bhai bhaskar ojha 2nd term dekh 1/(n+1)(n+4) even ban jayga galat kiya h tumne
Tn hoga= 1/(2n+1)*(2n+3)(2n+5)
Aur summation n=0 se start krna padega
14:45 sum of infinity terms is 1/3 and Sn Is 1/3-1/(2n+1)(2n+3)
Can Any one confirm
bro bahar 1/4 bhi rahega
Therefore, S (infinity) = 1/12
Bro you missed a factor of 1/4 .
Sinfinty is 1/12
You missed the factor of 1/4
yes I Missed 1/4 factor. Thanks for reminding
S(infinite)=11/24
27:10
Ans =1
🙂 it's the correct answer
If any one needs solution reply karna I'll tell you
I got the n term but no idea how to solve further
So please help
😅😅
Kya majak krre ho fraction me answer ayega dekhke hi lag rha h
@@pranavpurohit4693 la bhai tu fr answer maine ye question VMC illuminati sheet mai kiya hai waha 1 hi dia hai tu Le a apne hisab se fraction mai amswer maine kab mana kiya hao
Bata Bhai kaise kiya
Bro how did u get the answer then?
S infinite=11/24
Why did i not find this channel before 🥲?
Us bro us🥲
beocz ur a legend..by name
Cuz you are a legend both by your name and profile pic.....
Also you must be very good at piano and calligraphy....and maybe tea ceremony.
Sn = 2^2/3 - 1 for type 7
bro ☠
❤❤