# 9 - Binomial Theorem Questions Answers

Maths >> Algebra >> Binomial Theorem Engineering Exam

Find sum of series..1*(n)2  + 2*(n-1)+3*(n-2)2 +...+n .

Joshi sir comment

general term of the series = r(n+1-r)= r(n+1)2 + r3 - 2(n+1)r2

now put sigma then solve

If x+y = 1 ,then ∑ r  nCrxryn-requals nx....How??

Joshi sir comment

∑ r nCrxryn-r

∑ r nCrxr(1-x)n-r

∑ r n!/r!(n-r)xr(1-x)n-r

= n (n-1)!/(r-1)!(n-r)! xr(1-x)n-r

= n (n-1)C(r-1) xr(1-x)n-r

= n (n-1)Cxr+1(1-x)n-r-1             let r = r+1

= nx  (n-1)Cxr(1-x)n-r-1

now solve

What is the value of 0⁄0? if its otherthan 1, then how this can be true:      lim      (ex-1)/x = 1

x −−>0

Joshi sir comment

exact 0 and lim tends to 0 are different

Maths >> Algebra >> Binomial Theorem Board Exam

what is the range of    log 203?

Joshi sir comment

There should be any variabe in the question

find the value of [(1+0.0001)10000] WHERE [.] REPRESENTS GREATEST INTEGER FONCTION.

Joshi sir comment

(1 + 0.0001) 10000 = 1 + 10000*0.0001 + other terms = greater then 2

besides it lim x->0 (1+x)1/x = e = less then 3

it means the given expression is more then 2 but less then 3

so 2 will be the answer

consider 5 american couples and 2 indian couples sitting beside each other in n2 tables in the form of log5x.

Joshi sir comment

The coefficient of λ^n μ^n in the expansion of [(1+λ)(1+μ)(λ+μ)]^n is ?

Joshi sir comment

(1+λ)n = C0 + C1x + ..................................Cnxn

similarly for (1+μ)n and

(λ+μ)n = C0λn + C1λn-1μ + ........................ + Cnμn

On multiplying these 3, we will get the coefficient of  λnμ= ∑Cn3      (Here you should remember that C0 = Cn, C1 = Cn-1 and etc)

For getting the value of  ∑Cn:

multiply the expansions of (1+x)n, (1-x)n and [1-(1/x2)]n and calculate the coefficient of x0, we get ∑Cn3

and if we multiply (1+x)n, (1-x)n and [1-(1/x2)]n, then we get (-1)n(1-x2)2n/x2n so coefficient of xin this expression will be (-1)n 2nC(-1)2nCn

prove that the integral part of binomial expention is even
(5√5 + 11)2n+1

Joshi sir comment

Let x = (5√5 + 11)2n+1 and y = (5√5 - 11)2n+1

so x*y = 42n+1 = even number

similarly x - y = integer

if we consider the value of y , it will be (5*2.3-5)2n+1 = (0.5)2n+1  = fraction number less than 1   (approx)

so definitely y will be the fractional part of x, and difference of x and y is even so integer part of x will be even

The number of real negative terms in the binomial expansion of (1+ix)(4n-2),  n  N and x > 0 is?