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Past Year Questions › All exams › QA › Binomial Theorem

Binomial Theorem — PYPs

2 solved Binomial Theorem previous year questions (PYQs) from past year papers — attempt each and check the answer.

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Q1:ipmat indore 2023QA › Binomial TheoremMediumSA · TITA
If three consecutive coefficients in the expansion of (x+y)n(x+y)^n are in the ratio 1:9:631:9:63, then the value of nn is
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The Setup: We are given three consecutive coefficients in the binomial expansion of (x+y)n(x+y)^n strictly in the ratio 1:9:631:9:63. We must determine the power nn.

Step 1: Express the coefficients algebraically.
Let the three consecutive coefficients be
(nr−1)\binom{n}{r-1}, (nr)\binom{n}{r}, and (nr+1)\binom{n}{r+1}.
We are given two ratio equations:

(nr)(nr−1)=91=9\frac{\binom{n}{r}}{\binom{n}{r-1}} = \frac{9}{1} = 9

(nr+1)(nr)=639=7\frac{\binom{n}{r+1}}{\binom{n}{r}} = \frac{63}{9} = 7

Step 2: Apply the standard binomial coefficient ratio formula.
The ratio
(nk)(nk−1)\frac{\binom{n}{k}}{\binom{n}{k-1}} simplifies universally to n−k+1k\frac{n - k + 1}{k}.
Apply this to our first equation (
k=rk = r):

n−r+1r=9  ⟹  n−r+1=9r  ⟹  n=10r−1\frac{n - r + 1}{r} = 9 \implies n - r + 1 = 9r \implies n = 10r - 1

Apply this to our second equation (k=r+1k = r + 1):

n−(r+1)+1r+1=7  ⟹  n−rr+1=7  ⟹  n−r=7r+7  ⟹  n=8r+7\frac{n - (r+1) + 1}{r+1} = 7 \implies \frac{n - r}{r + 1} = 7 \implies n - r = 7r + 7 \implies n = 8r + 7

Step 3: Solve the linear system for rr and nn.
Equate the two expressions for
nn:

10r−1=8r+710r - 1 = 8r + 7

2r=8  ⟹  r=42r = 8 \implies r = 4

Substitute rr back into either equation to find nn:

n=10(4)−1=39n = 10(4) - 1 = 39

Final Answer: 39

Q2:ipmat indore 2022QA › Binomial TheoremEasySA · TITA
The sum of the coefficients of all the terms in the expansion of (5x−9)4(5 x-9)^{4} is __________.
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The Setup: We are asked to find the arithmetic sum of the coefficients of all terms within the binomial expansion of (5x−9)4(5x-9)^4.

Step 1: Apply the polynomial coefficient property.
For any polynomial
P(x)P(x), the sum of its fully expanded coefficients is mathematically obtained by evaluating the polynomial at x=1x = 1.

Step 2: Evaluate the expression at x=1x = 1.

P(1)=(5(1)−9)4P(1) = (5(1) - 9)^4

P(1)=(5−9)4P(1) = (5 - 9)^4

P(1)=(−4)4P(1) = (-4)^4

P(1)=256P(1) = 256

Final Answer: 256

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