# BINOMIAL EXPANSION PRACTICE QUESTIONS

Binomial Expansion Practice Questions :

Here we are going to see how to find expansion using binomial theorem.

Question 1 :

If a and b are distinct integers, prove that a − b is a factor of an − bn, whenever n is a positive integer. [Hint: write an = (a − b + b)n and expand]

Solution :

an  =  [(a - b) + b]n

annC0(a-b)n+nC1(a-b)n-1b1+nC2(a-b)n-2b2+nCn-1(a-b)bn-1+ nCnbn

an - bn  = (a-b)n+nC1(a-b)n-1b1+nC2(a-b)n-2b2+nCn-1(a-b)bn-1

an - bn

= (a-b) { (a-b)n-1 nC1(a-b)n-2b + nC2(a-b)n-3bnCn-1bn-1 }

Hence a − b is a factor of an − bn, whenever n is a  positive integer.

Question 2 :

In the binomial expansion of (a + b)n, the coefficients of the 4th and 13th terms are equal to each other, find n.

Solution :

The coefficients of the fourth ad thirteenth terms in the binomial expansion of (a + b)n are nC3 and nC12 respectively.

Coefficient of 4th term in (a+b)n = coefficient of 13th term in (a + b)n

nC3  =  nC12

If nCx  =  nCy   ==>  x = y or x + y = n

n  =  15

Question 3 :

If the binomial coefficients of three consecutive terms in the expansion of (a + x)n are in the ratio 1 : 7 : 42, then find n.

Solution :

Let the three consecutive terms be rth, (r+1)th and (r+2)th terms.Their coefficients in the expansion of (1+x)n are nCr-1, nCr and nCr+1 respectively. It is given that,

nCr-1 nCr : nCr+1  = 1  :  7  :  42

nCr-1 / nC =  1 / 7

r/(n-r+1)  =  1/7

n - 8r + 1 = 0  ----(1)

nCr : nCr+1  =  7/42

r+1/n-r  =  1/6

n - 7r - 6  =  0  -----(2)

(1) - (2)  ==>  (n - 8r + 1) - (n - 7r - 6)  =  0

n - n - 8r + 7r + 1 + 6  =  0

-r + 7  =  0  ==> r  =  7

By applying the value of r in the 1st equation, we get

n - 8(7) + 1 = 0

n - 56 + 1  =  0

n  =  55

Hence the values of n and r are 55 and 7 respectively.

Question 4 :

In the binomial coefficients of (1 + x)n, the coefficients of the 5th, 6th and 7th terms are in AP. Find all values of n.

Solution :

The coefficients of 5th, 6th and 7th terms in the binomial expansion of (1+x)n are nC4, nC5 and nC6 respectively,We are given that

2 nCnC4 + nC6

2  =  (nC4)/(nC5)  + (nC6)/(nC5)

2  =  5/(n-4) + (n-5)/6

2  =  [30 + (n-4)(n-5)]/[6(n-4)]

12(n-4)  =  =  [30 + (n2 - 4n - 5n + 20)]

12n - 48  =  30 + n2 - 9n + 20

n2 - 9n + 20 - 12n + 48 + 30  =  0

n2 - 21n + 98  =  0

(n - 14) (n - 7)  =  0

n - 14 = 0   n - 7 = 0

n  =  14,  n  =  7

Question 5 :

Prove that

Solution :

n C =  C0 , C 1  =  C1 ,...............

L.H.S

=  (C0)2+ (C1)+ (C2)2 + .............  + (Cn)2

=  (nC0)2 + (nC1)2 + (nC2)2 + .................. + (n C n)2

=  (nC0)(nC0) + (nC1)(nC1) + (nC2)(nC2) + ................. (nCn)(nCn)

=  (nC 0)(nCn-0) + (nC1)(nCn-1) + (nC2)(nCn-2) + ................. (nCn)(nCn-n)   -----(A)

(1 + x) =  nC0 1x+ nC1 1n-1 x1 + nC2 1n-

x2 + ..........+ nCn-1 1n-1 xn-1 + nCn 1n-nxn

(1 + x) =  nC0 + nC1 x1 + nC2 x2 + ..........+ nCn-1 xn-1 + nCn x ----(1)

By writing the expansion in reverse order, we get

(1 + x) =  nCx+ nCn-1 xn-1+ ..........+ nCx1  + nC0---(2)

(1) ⋅ (2)

(1 + x)2n  =  nCnC+ nCnCn-1 + ..................... + nC1 nCn-1 + nCnC----(B)

Since (A) and (B) are equal, we may consider the simplified value of L.H.S is (1 + x)2n

Tr + 1   =   n C r xn-r ar

x  =  1, a  =  x and n  =  2n

Tr + 1   =   2n C r 12n-r xr

here r  =  n

Tn + 1   =   2n C n 12n-n xn

=  2n!/n(2n-n)! 1n xn

Coefficient of xn  =  (2n)!/(n!)2

Hence proved.

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