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Syllabus

Find the term independent of x in the expansion of (2x - 1/x)

^{10}T?_{r+1}=^{n}Cr .a^{n-r} .b^{r}And when should you use this formula-

T?_{r+1}= (-1)^{n}C_{r }.a^{n-r}.(+b)^{r}the second, third and fourth term of the binomial expansion (x+a)n (n is actually (x+a)raised to the power n) are 240, 720 and 1080. find x, a and n.

Expand the Binomial (1-3x)

^{5}If the coeffients of 5th, 6th & 7th terms in expansion of (1+x)

^{n}are in AP, then find values of n???the coefficient of x

^{4}in the expansion of (1+x+x^{2}+x^{3})^{11}is :a) 900 b)909

c) 990 d)999

If 3rd,4th,5th,6th term in the expansion of (x+alpha)

^{n}be respectively a,b,c and d, prove that b^{2}-ac/c^{2}-bd=4a/3c..Show that C_{0}/2 + C_{1}/3 + C_{2}/4 + ......... + C_{n}/n+2 = (1+n.2^{(n+1)})/(n+1)(n+2)Please tell me the answer to this question. Need urgently. Help from meritnation experts would be commendable . Please help !

if 4th term in the expansion of ( ax+1/x)

^{n }is 5/2, then the values of a and n :a) 1/2,6 b) 1,3

c) 1/2,3

in the second example, just by finding n2=9, we can conclude that n=3. so why did we have to find n3=27?

The coefficients of three consecutive terms in the expansion of(1+x)

^{n}are in the ratio 1:7:42. find n.show that the coefficient of the middle term in the expansion of (1+x)^2n is equal to the sum of the coefficients of two middle terms in the expansion (1+x)^2n-1.

Show that the middle term in the expansion of(1+x)raise to power 2n is = 1.3.5.......(2n-1) . 2n.xraise to power n upon n! , where nis a +ve integer.

Using Binomial theoram, prove that 2

^{3n }- 7n^{}-1 is divisible by 49 where n is a Natural number^{2}+1/x]^{12}solve this

if the coefficients of (r-5)

^{th}and (2r-1)^{th}term in the expansion of (1+x)^{34}are equal, fiind r^{n}C_{0}+^{n}C_{1}+^{n}C_{2}+...........+^{n}C_{n}= 2^{n}(1+2x+x^2)^20

Using binomial theoram ,show that 9

^{n+1}-8n-9 is divisible by 64 ,whr n is a positive integer.^{3})((3/2)x^{2}- 1/3x)^{9.}7. If the coefficient of r

^{th},(r+1)^{th}and (r+12^{th}terms in the binomial expansion of (1+y)^{m}are in AP, then m and r satisfy the equation:(1) m

^{2}-m(4r-1)+4r^{2}+2=0 (2) m^{2}-m(4r+1)+4r^{2}-2=0(3) m

^{2}-m(4r+1)+4r^{2}+2=0 (4) m^{2}-m(4r-1)+4r^{2}-2=08. The Value of ${C}_{4}^{50}+\sum _{r=1}^{6}{C}_{3}^{56-1}$ is

$\left(1\right){C}_{4}^{56}\left(2\right){C}_{3}^{56}\left(3\right){C}_{3}^{55}\left(4\right){C}_{4}^{55}$

using binomial therorem, 3

^{2n+2}-8n-9 is divisible by 64, n belongs to NFind the value of nC0 - nC1 + nC2 - nC3 +.................+(-1)^n nCn

prove nPr=

n!(n-r)!

and deduce it to nCr answere fast important for test

if three successive coefficients in the expressions of (1+x)

^{n}are 220, 495 and 792 respectively, find the value of n?Find

a,bandnin the expansion of (a+b)^{n}if the first three terms of the expansion are 729, 7290 and 30375, respectively.If (1+x)

^{n}= C_{0}+ C_{1}x + C_{2}x^{2}+ …..+C_{n}x^{n}. Prove that C_{1}+ 2C_{2}+ 3C_{3}+ ……+ nC_{n}= n.2^{n-1 }Find

n, if the ratio of the fifth term from the beginning to the fifth term from the end in the expansion ofprove that

^{n}C_{r}+^{n}C_{r-1}=^{n+1}C_{r}Find the sixth term of the expansion (y

^{1/2}+ x^{1/3})^{n}, if the binomial coefficient of the third term from the end is 45.the first three terms in the expansion of (x+y)^n are 1,56,1372 respectively.Find x and y

^{2}-x^{3}/6)^{7}^{10}in the copy^{40}in the expansion of (1/x^{2}+ x^{4})^{18}^{st}4 terms in the expansion of ( 1 –x )^{-1/4}^{2}/4)^{9}^{3}The cofficient of three consecutive terms in the expansion of (1+x)

^{n}are in the ratio 1:7:42.find n?^{10}^{-17 }on the expansion of (x^{4}-1/x^{3})^{15.}.^{1/3}+x^{-1/5)}^{8.}^{8}*y16 in the expansion of (x+y)^{18.}The sum of the coefficients of the first three terms in the expansion of (x-3/x. (NCERT PG 174 EXAMPLE NO 16). The steps in the NCERTbook are not clear..^{2})^{m}, x is not equal to 0,m being a natural number, is 559. Find the term of the wxpansion containing x^{3}The 3rd, 4th and 5th terms in the expansion of (x + a)n are respectively 84, 280 and 560, find the values of x, a and n.

how to find coeeficient of a

^{4 }in the product(1+2a)^{4}(2-a)^{5}the coefficients of 2nd, third and fourth terms in the expansion of (1+n)^2n are in AP.Prove that 2n^2-9n+7=0

The no of terms with integral coefficients in the expansion (17

^{1/3 }+ 35^{1/2x})^{600 }is?????(Ans 101)if C1, C2, c3, C4 are the coefficient of 2nd, 3rd , 4th , and 5th , term in the expansion of (1+x)raise to "n" then prove that

C1/C1+c2 + c3/c3+c4 = 2c2/c2+c3 ((((((((((((((((((((( where c1+c2 , c3 +c4 and c2 + c3 are together under the division THAT IS C1 BY C1 +C2 etc.))))))))))))

In the Q14 NCERT TB.Ex: 8.1. In the discussion part an alternative method was postd. Ans: expand L.H.S. =

^{n}C_{0}+^{n}C_{1}.3 +^{ n}C_{2}.3^{2}+^{n}C_{3}.3^{3}+ ............. +^{n}C_{n}.3^{n}=(1+3)

^{n}=4^{n}=R.H.S.Hence Proved. Please explain how?

In the binomial expansion of( a - b )

^{n}, if n â‰ 5 and the sum of 5^{th}and 6^{th}terms is 0, then finda/bin the binomial expansion of (a + b)

^{n}, the coefficient of the 4th and the 13th terms are equal to each other. find n?if the coefficient of (r-1)th, rth and (r+1)th terms in the expansion of (1+x)^n are in the ratio 1:7:42, find n and r

^{2})^{4}for questions such as to find that which is greater of (1.01)^1000000 or 10000

in ncert solution it is given that

^{1000000C}_{1}=1000000 BUT atq formula it shd hv been 1000000!/99999!*1pls help why??? reply soon :/

_{}_{}_{}_{}This is my doubt:

Find a if the coefficients of x

^{2}and x^{3}in the expansion of (3+ax)^{9 }are equal.Thanks a lot. =)

^{10}The sum of two numbers is 6 times their geometric mean show that the numbers are in the ratio

(3+2.2^{1/2}):(3-2.2^{1/2})_{1}/C_{0}) + (2C_{2}/C_{1}) + ( 3C_{3}/ C_{2}) +.... + nC_{n}/C_{n-1}= ? Pls solve using summation method. Thanks^{10}C_{r}^{20}C_{m-r}is maximum when:Show that 2

^{4n}-15n-1 is divisible by 225 by using binomial theorem.Q4. Prove that : ${}_{}{}^{\mathrm{n}-1}\mathrm{C}_{\mathrm{r}}+{}_{}{}^{\mathrm{n}-2}\mathrm{C}_{\mathrm{r}}+{}_{}{}^{\mathrm{n}-3}\mathrm{C}_{\mathrm{r}}+......+{}^{\mathrm{r}}{\mathrm{C}}_{\mathrm{r}}{=}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}+1}$.

using binomial theorem prove that 6

^{n}-5n always remender -1when divided by 25Find the fifth term from the end in the expansion of (x

^{3}/2 - 2/x^{2})^{9}find the coefficient of x

^{n}in the expansion of(1+x)(1-x)^{n}any 3 successive coefficient in the expansion of (1+x)^n where n is a positive integer are 28,56,70 then n is

please give the blueprint of annual examination of maths paper.

SOLVE

1) C1+2C2+3C3+--------+nCn=n2 to power n-1

if the 21st and 22nd terms in the expansion of (1+x)^44 are equal then find the value of x.

The no of irrational terms in the expansion of (4

^{1/5}+ 7^{1/10})^{45 }are??????^{14}Find the coefficient of x

^{50 }in the expansion :(1+x)

^{1000}+ 2x(1+x)^{999}+3x^{2}(1+x)^{998}+…………………..+1001x^{1000}1. Find the total no. of terms in the expansion of (x+a)^100 + (x-a)^100after simplification

^{n}E_{r=0}^{n}C_{r}sin2rx /^{n}E_{r=0}^{n}C_{r}cos2rx = tan nxE--->Sigma(summation).

The first 3 terms in the expansion of (1+ax)

^{n}are 1, 12x, 64x^{2}respectively, Find n and 'a' .Find n if the ratio of the 5th term from the beginning to the 5th term from the end in the expansion of (4√2 + 1/4√3)

^{n }is √6:1?If the coefficient of x

^{r}in the expansion of (1-x)^{2n-1}is denoted by a_{r}then prove that a_{r-1}+ a_{2n-r}= 0.if (1+ax +bx

^{2})(1-2x)^{18}be expressed in ascending powers of x such that coefficient of x^{3}and x^{4}are 0 then (a,b)is equal to :a) (16,272/3) b) (4,44/3)

c)(1/16,88/3)