Sir please do complex numbers next. Especially geometric application of complex numbers. Great content as usual thank you for your work.
@anirudhsrivastava6042
2 жыл бұрын
Yes i agree
@anshumanarya9215
2 жыл бұрын
Yess please 🙏
@krishna_000
2 жыл бұрын
Yes please
@i.am.harsh.
2 жыл бұрын
Yes.
@friendlyneighbourhoodbrodude
2 жыл бұрын
Yes agreed.
@krishna_000
2 жыл бұрын
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.
@adityaagrawal5407
2 жыл бұрын
u r a1
@aditsaini5094
Жыл бұрын
@@adityaagrawal5407 a100
@krishsharma7088
2 жыл бұрын
Type 3 : S infinity = 1/12
@Jee_anthem
8 ай бұрын
-1/12
@shubhsoni5959
8 ай бұрын
Yes bro my also same
@shubhsoni5959
8 ай бұрын
@@Jee_anthem no its 1/12 maam
@lakshay3745
7 ай бұрын
Galt bhi confidence se bolte ho😂@@shubhsoni5959
@saurabhrana4160
2 жыл бұрын
Type 6 is a question of the IOQM Examination. Beautiful indeed!
@su-bro
2 жыл бұрын
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
@wanwanmlbb7100
Жыл бұрын
We know this shit
@kshitijverma7238
2 жыл бұрын
been preparing for jee from last 3 1/2 years how the hell i never found this channel before 😭😭
@tirtheshjadhav1898
2 ай бұрын
3 and a half years? Damn
@navalgurjar2070
2 жыл бұрын
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
@shivankulcodezz
2 жыл бұрын
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
@conceptisthekey5643
2 жыл бұрын
But if diff becomes GP then this leads to wrong ans.
@amoghsrivastava6324
2 жыл бұрын
thank you sir but have a regret for not exploring this channel before
@RZMATHS
2 жыл бұрын
Thank you so much .... Please like and share the video so that many students get the benefit out of these videos.
@amoghsrivastava6324
2 жыл бұрын
@@RZMATHS already shared in all my coaching groups and all extra channels :)
@shreyanshmishra6971
2 жыл бұрын
@@amoghsrivastava6324 unacademy....?
@amoghsrivastava6324
2 жыл бұрын
@@shreyanshmishra6971 yes and are you albatross ??
@jeeadvanced10
2 жыл бұрын
This was the holy pill of wisdom i needed to solve questions of series summations. Thankyou sir🙂
@abhinav788
2 жыл бұрын
Type - 3 H.W. -- S(infinity)= 1/12
@vanshajverma2199
2 жыл бұрын
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
@shubhamgupta-4425
2 жыл бұрын
Type 6 was literally unknown for me.thanku sir for giving and solving Advance problems .
@yadav_nitish428
3 ай бұрын
Thanks a lot sir just cleared all cocnepts with ques in just a short time...
@chromeee_7421
2 жыл бұрын
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
@farista4667
2 жыл бұрын
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
@himeshkumar5293
2 жыл бұрын
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
@adityashriwal5378
2 жыл бұрын
@@himeshkumar5293 I have also form same rth term like u but what to do next???
@nomadgamercao1177
2 жыл бұрын
@@himeshkumar5293 right bro
@himeshkumar5293
2 жыл бұрын
@@adityashriwal5378 right 1 as 2n+2-2n-1 in numerator
@adityashriwal5378
2 жыл бұрын
@@himeshkumar5293 ok
@himanshusharma4268
2 жыл бұрын
such a fantastic lect on sum of series for jee adv
@temochnu
2 жыл бұрын
I am got admire of u sir glad to see a wonderful mathematician sir
@cutegirlakanshabag2449
2 жыл бұрын
Thank you so much sir.. 😊😊👍👍Type -6 question 10MQ h 😊👍
@tommyshelby6277
2 жыл бұрын
sir pls next one on Matrices to be specific: questions on adjoint, orthogonal matrix, inverse matrix
@ravibaraskar6007
2 жыл бұрын
thanks sir aap probability or application of derivatives ke bhi tricks and advance problems upload kijiye ..
@atultiwari9528
2 жыл бұрын
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
@anurag_gautam00
2 жыл бұрын
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 🙏
@jatinreddy
10 ай бұрын
Type 5: exact question was asked in Mains 1st Feb Morning shift 2023
@sanjaijee7157
2 жыл бұрын
Please DO Complex Numbers or3D GEO. on upcoming videos. I loved and learned a lot sir
@Volase
2 ай бұрын
This was extremely helpful
@PIYUSH777MyNameIsPiyushSoni
8 ай бұрын
Sir pls cover the factorial summation aswell
@m.rohithkumar8212
2 жыл бұрын
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..
@vivek-hb1ht
2 жыл бұрын
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)]
@kushaankoul5775
6 ай бұрын
whats the general term
@shyger778
2 жыл бұрын
loving ur videos sir thanks a bunch :)
@RZMATHS
2 жыл бұрын
Thank you so much ❤️... Please share the videos with your friends too...☺️
@shyger778
2 жыл бұрын
@@RZMATHS 😇
@Mysterious-735
2 жыл бұрын
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
@devanshukumar5740
2 жыл бұрын
Sir please make a lecture on trigonometric summation series
@rahulkumarsingh4137
2 жыл бұрын
this was the same video i was wanting, you did the same.
@suhassutre4658
2 жыл бұрын
best explanation sir thank u for such efforts
@akshatkarkar6588
2 жыл бұрын
Thanks sir.....very good videos 👍
@OCEAN_NINJA
5 ай бұрын
you are ammazin I enjoy a lot
@herlim6927
2 жыл бұрын
1/(2r-1)(2r+1)(2r+3)
@perfect5660
2 жыл бұрын
Type 6 was good👍
@GulshanJii-rw5dm
4 ай бұрын
In type 1 sum of first n natural no. Is n(n+1)/2
@nishantrathore5031
2 жыл бұрын
Thank you sir for helping for us
@Abhi-kr6df
2 жыл бұрын
27:37
@hiteshverma1786
2 жыл бұрын
S infinity=1/12
@shubhsoni5959
8 ай бұрын
Thank you very much sirji
@shivamshukla9129
3 ай бұрын
You are great sir❤
@vishalmishra0014
2 жыл бұрын
Sn = 2^2/3 - 1 for type 7
@kushaankoul5775
6 ай бұрын
bro ☠
@commonman248
Ай бұрын
Thank you sir please make video on trigo
@a-vinil
2 жыл бұрын
Sir, please teach how to factorise the denominator as you did at 20:49
@RZMATHS
2 жыл бұрын
Check my inverse trigonometric series revision i have done the same factorisation there.
@a-vinil
2 жыл бұрын
@@RZMATHS ok
@a-vinil
2 жыл бұрын
@@RZMATHS Sir, I couldn't find it in all the trignometric series videos. Could you please make a video on that.
@uditraj5729
2 жыл бұрын
@@a-vinil Clearly sir said it's in inverse trigonometric series video
@riteshpatil1941
4 ай бұрын
❤❤
@gagannarang4362
2 жыл бұрын
BINOMIAL COEFFICIENT PART 2 SIRR SOOON PLEASE
@manamtiwari
2 жыл бұрын
Was waiting for these kind , awesome sir
@prtk2verma
2 жыл бұрын
Sir take a session on question solving approach in AOD (highest weightage topic of adv.)
@savagesyd29
2 жыл бұрын
Thanks for the quick revision
@itsmathslover...6617
2 жыл бұрын
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 ??
@jeevan3226
2 жыл бұрын
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
@arvind_24
2 жыл бұрын
Thank you sir ❤❤❤
@yashshindekar9088
2 жыл бұрын
Sir please do continue this IIT Jee series likely to come for Jee advance 2023.
@ruhailkhan3165
2 жыл бұрын
S of inf= 1/12 15:00
@parijatgoyaliitkgp7572
2 жыл бұрын
Type 7, S(inf)=1
@manas-singh
2 жыл бұрын
Sir its humble request aap compound angles fully jee adv++ level pr explain kr dijiye
@shailendrasingh9873
2 жыл бұрын
Btw apart from everything,sir why is this channel named rz maths ? Is the shortform of your name is rz?
@RZMATHS
2 жыл бұрын
Yup
@goodtuber3411
Жыл бұрын
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...🙏
@AnkitSingh-mv1oi
2 жыл бұрын
Loved it 😊
@olliver8031
2 жыл бұрын
Sir please upload a video on jee advanced problem on 2 newton lebinitz formula Rollle's theorem
@monurathore3241
2 жыл бұрын
Thankyou bhaiya
@Name-mj9rx
2 жыл бұрын
Intial concept used in limits
@soumyanamdeo2347
2 жыл бұрын
sir please next video on trigonometric series 🙏🙏
@geetapandey9956
2 жыл бұрын
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
@Trueneutron
2 жыл бұрын
Thanks a lot
@atharvasheershpandey2097
Жыл бұрын
Pretty GOOD
@aditya1dhiman
2 жыл бұрын
these are very helpful sir🙏
@himanshugurjar5630
2 жыл бұрын
s infinity of last qustion is 1
@gauravahuja7803
2 жыл бұрын
Thx sir
@divyanshgupta2012
2 жыл бұрын
Sir 0lease do vector 3d next
@indreshpatel7764
2 жыл бұрын
SIR PLEASE COMPLETE BINOMIAL SERIES LEFT PART
@francisangelmanohar5020
2 жыл бұрын
Thank you sir
@yashs.gosavi4024
2 жыл бұрын
Why did i not find this channel before 🥲?
@ayyamguari6134
2 жыл бұрын
Us bro us🥲
@tinywrestler383
2 жыл бұрын
beocz ur a legend..by name
@mehulgoyal7727
2 жыл бұрын
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.
@vikask2512
2 жыл бұрын
Complex Numbers Please Sir.....
@xeoldboy9714
Жыл бұрын
Helpfull
@sudhirtiwary8111
2 жыл бұрын
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
@geetapandey9956
2 жыл бұрын
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
@parijatgoyaliitkgp7572
2 жыл бұрын
Sir when is the second part of binomial lecture coming ?
@ligma_hairy-ass
2 жыл бұрын
We need part 2
@imankalyanbhunia1263
2 жыл бұрын
Sir please coordinate geometry question series
@ruhailkhan3165
2 жыл бұрын
Sir can you make vidoes on limits?
@khushpatel8176
2 жыл бұрын
Can u pls as well add link to pdf notes of these videos
@ariflear_reactor
2 жыл бұрын
Sir please make a leacture in hindi
@kiruthika4671
2 жыл бұрын
After seeing your thumbnail I m freaking out to leave jee advanced
@4zvm
2 жыл бұрын
Lastone 💥
@ARYAN_BAJPAI_IIT_BHU
2 жыл бұрын
Tele gram group banao uspe discussion Kiya jye
@yashpandey3.3.3
2 жыл бұрын
You are missing out matrices and dets.
@AMANKUMAR-ww9us
2 жыл бұрын
15:00 answer for S(infinity) is 1/60.
@AMANKUMAR-ww9us
2 жыл бұрын
Thanks bhai
@pranavpurohit4693
2 жыл бұрын
@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
@anaysingh-iitdelhi8955
2 жыл бұрын
27:10 Ans =1 🙂 it's the correct answer If any one needs solution reply karna I'll tell you
@lakshya254
2 жыл бұрын
I got the n term but no idea how to solve further So please help 😅😅
@pranavpurohit4693
2 жыл бұрын
Kya majak krre ho fraction me answer ayega dekhke hi lag rha h
@anaysingh-iitdelhi8955
2 жыл бұрын
@@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
@abuzarazhar7959
2 жыл бұрын
Bata Bhai kaise kiya
@sabhyatachoudhary5491
3 ай бұрын
Bro how did u get the answer then?
@vempatisridhar7448
2 жыл бұрын
14:45 sum of infinity terms is 1/3 and Sn Is 1/3-1/(2n+1)(2n+3) Can Any one confirm
@samarthchaplot2998
2 жыл бұрын
bro bahar 1/4 bhi rahega Therefore, S (infinity) = 1/12
Пікірлер: 152