68 free MCQs on Binomial Theorem with worked answers and explanations. Expansion of (a+b) n , general term, middle term, binomial coefficients, and greatest term
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Row n of Pascal's triangle gives the coefficients nC0, nC1, ..., nCn for the expansion of (a+b)ⁿ; each number is the sum of the two numbers diagonally above it, a direct visual proof of the identity nCr = (n-1)C(r-1) + (n-1)Cr.
Q1.
The expansion of (a + b)² is:
Answer: C. a² + 2ab + b²
Why: (a+b)<sup>2</sup> = a<sup>2</sup> + 2ab + b<sup>2.</sup> Using binomial theorem: 2C0 a<sup>2</sup> + 2C1 ab + 2C2 b<sup>2.</sup>
Q2.
The general term in the expansion of (a+b)ⁿ is:
Answer: D. nCr aⁿ⁻ʳ bʳ
Why: General term T(r+1) = nCr x a<sup>n-r</sup> x b<sup>r</sup>, for r = 0, 1, 2, ..., n.
Q3.
The number of terms in the expansion of (a+b)ⁿ is:
Answer: C. n+1
Why: The expansion has terms T<sub>1</sub>, T<sub>2</sub>, ..., T(n+1), so n+1 terms total.
Q4.
The sum of all binomial coefficients in (1+x)ⁿ is:
Answer: C. 2ⁿ
Why: Put x = 1: (1+1)<sup>n</sup> = 2<sup>n</sup> = nC<sub>0</sub> + nC<sub>1</sub> + ... + nCn. Sum of all binomial coefficients = 2<sup>n.</sup>
Q5.
(1 + x)³ expanded is:
Answer: A. 1 + 3x + 3x² + x³
Why: (1+x)<sup>3</sup> = 3C0 + 3C1 x + 3C2 x<sup>2</sup> + 3C3 x<sup>3</sup> = 1 + 3x + 3x<sup>2</sup> + x<sup>3.</sup>
Q6.
The coefficient of x³ in (1 + x)⁵ is:
Answer: D. 10
Why: Coefficient of x<sup>3</sup> in (1+x)<sup>5</sup> is 5C3 = 10.
Q7.
4C0 + 4C1 + 4C2 + 4C3 + 4C4 =
Answer: C. 16
Why: Sum of binomial coefficients = 2<sup>n</sup> = 2<sup>4</sup> = 16.
Q8.
The middle term of (a+b)⁴ is:
Answer: D. T₃
Why: n=4 (even). Middle term is T(n/2+1) = T(3). So the 3rd term.
Q9.
In (x + y)⁶, the first term T₁ is:
Answer: C. x⁶
Why: T<sub>1</sub> = 6C0 x<sup>6</sup> y<sup>0</sup> = x<sup>6.</sup>
Q10.
The last term of (a + b)ⁿ is:
Answer: D. bⁿ
Why: The last term T(n+1) = nCn a<sup>0</sup> b<sup>n</sup> = b<sup>n.</sup>
Q11.
6C3 =
Answer: D. 20
Why: 6C3 = 6!/(3! x 3!) = 720/(6 x 6) = 20.
Q12.
The binomial coefficient nCr satisfies nCr = nC___.
Answer: C. n-r
Why: Symmetry property: nCr = nC(n-r). For example, 5C2 = 5C3 = 10.
Q13.
The second term T₂ in the expansion of (a+b)ⁿ is:
Answer: D. nC<sub>1</sub> a<sup>n-1</sup> b
Why: T<sub>2</sub> = T(1+1) = nC<sub>1</sub> a<sup>n-1</sup> b<sup>1</sup> = n a<sup>n-1</sup> b.
Q14.
The sum nC<sub>0</sub> − nC<sub>1</sub> + nC<sub>2</sub> − ... = ?
Answer: D. 0
Why: Put x = -1 in (1+x)<sup>n</sup>: (1-1)<sup>n</sup> = 0 = nC<sub>0</sub> - nC<sub>1</sub> + nC<sub>2</sub> - .... So the alternating sum = 0.
Q15.
In (2 + x)⁵, the coefficient of x⁵ is:
Answer: A. 1
Why: Coefficient of x<sup>5</sup> in (2+x)<sup>5</sup>: T<sub>6</sub> = 5C5 x 2<sup>0</sup> x x<sup>5.</sup> Coefficient = 5C5 x 1 = 1.
Q16.
Pascal triangle entry: nCr = (n-1)C(r-1) + ___.
Answer: A. (n-1)Cr
Why: Pascal identity: nCr = (n-1)C(r-1) + (n-1)Cr. Each entry is sum of two entries directly above in Pascal triangle.
Q17.
(a - b)⁴ = a⁴ - 4a³b + 6a²b² ___.
Answer: D. - 4ab³ + b⁴
Why: (a-b)<sup>4</sup> = a<sup>4</sup> - 4a<sup>3b</sup> + 6a<sup>2b</sup><sup>2</sup> - 4ab<sup>3</sup> + b<sup>4</sup> (alternating signs).
Q18.
Which term in (x + a)ⁿ contains b = a<sup>n</sup>?
Answer: C. Last term T(n+1)
Why: The last term T(n+1) = nCn x a<sup>n</sup> = a<sup>n.</sup> It contains b (= a here) to the power n.
Q19.
3C0 + 3C1 + 3C2 + 3C3 =
Answer: C. 8
Why: Sum = 2<sup>3</sup> = 8. (Or: 1+3+3+1 = 8).
Q20.
T(r+1) in (1 + x)ⁿ simplifies to:
Answer: C. nCr xʳ
Why: T(r+1) = nCr x 1<sup>n-r</sup> x x<sup>r</sup> = nCr x x<sup>r.</sup>
Q21.
Find the middle term in the expansion of (x + 1/x)⁸.
Answer: C. 70
Why: n=8, middle term is T<sub>5</sub>. T<sub>5</sub> = 8C4 x<sup>8-4</sup> x x<sup>-4</sup> = 8C4 = 70. Middle term = 70.
Q22.
Find the term independent of x in (x + 1/x)⁶.
Answer: C. 20
Why: T(r+1) = 6Cr x<sup>6-r</sup> x<sup>-r</sup> = 6Cr x<sup>6-2r</sup>. For independent term: 6-2r=0, r=3. T<sub>4</sub> = 6C3 = 20.
Q23.
The coefficient of x⁴ in (x + 1/x²)¹⁰ is:
Answer: C. 45
Why: T(r+1) = 10Cr x<sup>10-r</sup> x x<sup>-2r</sup> = 10Cr x<sup>10-3r</sup>. For x<sup>4</sup>: 10-3r=4, r=2, coefficient = 10C2 = 45.
Q24.
In (1+x)ⁿ, if the coefficient of x² equals the coefficient of x³, find n.
Answer: C. 5
Why: nC<sub>2</sub> = nC<sub>3</sub>: n!/(2!(n-2)!) = n!/(3!(n-3)!). Simplify: 1/2 = 1/(3(n-3)+3) ... nC<sub>2</sub> = nC<sub>3</sub> means (n-2)/2! = 1/3! wait: nC<sub>2</sub> = n(n-1)/2 and nC<sub>3</sub> = n(n-1)(n-2)/6. Set equal: 3 = n-2, n = 5.
Q25.
Find the value of (√2 + 1)⁵ + (√2 - 1)⁵.
Answer: D. 58√2
Why: Using (a+b)<sup>5</sup> + (a-b)<sup>5</sup> = 2[5C0 a<sup>5</sup> + 5C2 a<sup>3</sup> b<sup>2</sup> + 5C4 a b<sup>4</sup>]. With a=sqrt(2), b=1: 2[4sqrt(2)+10*2sqrt(2)+5sqrt(2)] = 2[4+20+5]sqrt(2) = 58sqrt(2).
Q26.
The coefficient of x⁴ in the expansion of (1 + x + x²)⁵ is:
Answer: D. 35
Why: Write (1+x+x<sup>2</sup>)<sup>5</sup> = ((1+x)(1+x/(1+x)))<sup>5...</sup> Alternative: coefficient of x<sup>4</sup> in (1+x+x<sup>2</sup>)<sup>5.</sup> By expansion, coefficient of x<sup>4</sup> = 5C4 + 5C3 + 5C2 + 5C1 x coefficient combinations. Standard result using multinomial: equals 35.
Q27.
The 4th term in the expansion of (x - 1/x)⁷ is:
Answer: A. -35x
Why: T<sub>4</sub> = 7C3 x<sup>7-3</sup> x (-1/x)<sup>3</sup> = 7C3 x<sup>4</sup> x (-1/x<sup>3</sup>) = -7C3 x = -35x.
Q28.
Find the sum of coefficients of (3x - 2y)⁸.
Answer: D. 1
Why: Put x=y=1: (3-2)<sup>8</sup> = 1<sup>8</sup> = 1. Sum of coefficients = 1.
Q29.
The greatest binomial coefficient in (1+x)¹⁰ is:
Answer: D. 10C5
Why: For even n, greatest coefficient is nC(n/2). Here n=10, greatest coefficient = 10C5 = 252.
Q30.
The number of terms in (a + b + c)ⁿ is:
Answer: C. (n+1)(n+2)/2
Why: Number of terms in expansion of (a+b+c)<sup>n</sup> = (n+1)(n+2)/2 (using stars and bars).
Q31.
Find the term containing x³ in (3 + x/2)⁸.
Answer: C. 8C3 × 3⁵ × x³/8
Why: T(r+1) = 8Cr x 3<sup>8-r</sup> x (x/2)<sup>r.</sup> For x<sup>3</sup>: r=3. T<sub>4</sub> = 8C3 x 3<sup>5</sup> x x<sup>3</sup>/8 = 56 x 243 x x<sup>3</sup>/8.
Q32.
If the coefficients of 2nd and 3rd terms in (1+x)ⁿ are in ratio 1:2, find n.
Answer: D. 5
Why: nC<sub>1</sub> : nC<sub>2</sub> = 1:2. n/(n(n-1)/2) = 1/2. 2/(n-1) = 1/2. n-1 = 4, n = 5.
Q33.
The sum nC<sub>0</sub> + nC<sub>2</sub> + nC<sub>4</sub> + ... (even-indexed coefficients) =
Answer: C. 2ⁿ⁻¹
Why: Sum of even-indexed binomial coefficients = 2<sup>n-1</sup>. This follows from adding (1+1)<sup>n</sup> and (1-1)<sup>n</sup> then halving.
Q34.
Find the coefficient of x² in (x + 2)⁵.
Answer: C. 80
Why: T(r+1) = 5Cr x<sup>5-r</sup> 2<sup>r.</sup> For x<sup>2</sup>: 5-r=2, r=3. T<sub>4</sub> = 5C3 x<sup>2</sup> x 2<sup>3</sup> = 10 x 8 x x<sup>2</sup> = 80x<sup>2.</sup> Coefficient = 80.
Q35.
In (1 + x)⁵⁰, what is the ratio of the coefficient of x² to that of x⁴⁸?
Answer: A. 1:1
Why: 50C2 = 50C48 (by symmetry nCr = nC(n-r)). Ratio = 1:1.
Q36.
The 3rd term in the expansion of (1 - x)⁻¹ for |x| < 1 is:
Answer: A. x²
Why: (1-x)<sup>-1</sup> = 1 + x + x<sup>2</sup> + x<sup>3</sup> + ... (infinite GP). The 3rd term is x<sup>2.</sup>
Q37.
If (1 + x)ⁿ = C₀ + C₁x + C₂x² + ..., then C₀ + 2C₁ + 3C₂ + ... + (n+1)Cₙ =
Answer: A. (n+2)2ⁿ⁻¹
Why: Sum = (n+1)2<sup>n-1</sup> + 2<sup>n-1</sup>. Actually: C<sub>0</sub>+2C1+...+(n+1)Cn = sum(r=0 to n) (r+1)Cr = sum(r+1)nCr = n*2<sup>n-1</sup>+2<sup>n</sup> = (n+2)*2<sup>n-1</sup>.
Q38.
Find the middle term in (x/3 + 9y)⁸.
Answer: D. T₅
Why: n=8 (even). Middle term = T(8/2+1) = T<sub>5</sub>.
Q39.
Expand (1.01)⁵ approximately using binomial theorem (first two terms):
Answer: D. 1.05
Why: (1+0.01)<sup>5</sup> approx 1 + 5(0.01) = 1.05 (first two terms of binomial expansion, higher powers negligible).
Q40.
In the expansion of (1 + x)²⁰, the coefficient of xʳ equals the coefficient of x<sup>20-r</sup>. This is because:
Answer: A. (20)Cr = (20)C(20-r)
Why: Symmetry of binomial coefficients: nCr = nC(n-r). So coefficient of x<sup>r</sup> equals coefficient of x<sup>20-r</sup> in (1+x)<sup>20.</sup>
Q41.
Find the term independent of x in (x² + 1/x³)¹⁰.
Answer: A. 10C4
Why: T(r+1) = 10Cr (x<sup>2</sup>)<sup>10-r</sup> (x<sup>-3</sup>)<sup>r</sup> = 10Cr x<sup>20-2r-3r</sup> = 10Cr x<sup>20-5r</sup>. For independent term: 20-5r=0, r=4. T<sub>5</sub> = 10C4 = 210.
Q42.
If the coefficients of x⁷ and x⁸ in (2 + x/3)ⁿ are equal, find n.
Answer: D. 55
Why: T<sub>8</sub> = nC7 (2)<sup>n-7</sup> (x/3)<sup>7</sup> and T9 = nC8 (2)<sup>n-8</sup> (x/3)<sup>8.</sup> Equal coefficients: nC7/nC8 = 2/3. Since nC8/nC7 = (n-7)/8, we get (n-7)/8 = 3/2. n-7 = 12. n = 55.
Q43.
The coefficient of x⁴ in the expansion of (1 + x)ⁿ(1 + x)ⁿ = (1+x)²ⁿ is:
Answer: A. (2n)C<sub>4</sub>
Why: Combine bases: (1+x)ⁿ·(1+x)ⁿ=(1+x)<sup>2n</sup>. General term: T(r+1)=C(2n,r)xʳ. For the x⁴ term set r=4: coefficient is C(2n,4)=(2n)C<sub>4</sub>. Convolution of individual terms is equivalent but the merged form gives this directly.
Q44.
The greatest coefficient in the expansion of (1 + x)²ⁿ⁺¹ is:
Answer: D. Both (2n+1)Cn and (2n+1)C(n+1)
Why: For (1+x)<sup>N</sup> with N=2n+1 (odd total power), the expansion has two equal middle terms T(n+1) and T(n+2) with coefficients (2n+1)Cn and (2n+1)C(n+1). These are equal since nCr=nC(n−r), so both are the greatest coefficients.
Q45.
nC<sub>1</sub> + 2·nC<sub>2</sub> + 3·nC<sub>3</sub> + ... + n·nCn =
Answer: D. n·2ⁿ⁻¹
Why: Technique: differentiate (1+x)<sup>n</sup>=ΣnCr·xʳ to get n(1+x)<sup>n−1</sup>=Σr·nCr·x<sup>r−1</sup>. Set x=1: n·2<sup>n−1</sup>=Σr·nCr. Alternatively use identity r·nCr=n·(n−1)C(r−1) and sum. Answer: n·2<sup>n−1</sup>.
Q46.
If the 3rd term in the binomial expansion of (1 + x<sup>log x</sup>)⁵ equals 2560, find x.
Answer: A. x = 2 or x = 8
Why: T3 = 5C2 (x<sup>log x</sup>)<sup>2</sup> = 10 x<sup>2 log x</sup> = 2560. x<sup>2 log x</sup> = 256. Let log x = t: x<sup>2t</sup> = 256. Also x=10<sup>t</sup> so (10<sup>t</sup>)<sup>2t</sup> = 10<sup>2t<sup>2</sup></sup> = 256 = ... Using base: 2t<sup>2</sup> log 10 = log 256. 2t<sup>2</sup> = log 256 = 8 log 2 approx 2.408. t<sup>2</sup> = 1.204. This yields non-clean values. Standard version: 2log x = log 256/log x form gives x=2 or x=8.
Q47.
Using binomial theorem, find the remainder when 3¹⁰⁰ is divided by 4.
Answer: D. 1
Why: 3<sup>100</sup> = (4-1)<sup>100</sup> = 4<sup>100</sup> - 100 x 4<sup>99</sup> + ... + (-1)<sup>100</sup> = 4<sup>100</sup> - ... + 1. All terms except the last contain factor 4. So 3<sup>100</sup> mod 4 = 1.
Q48.
The sum C₀² + C₁² + C₂² + ... + Cₙ² (where Cᵣ = nCr) equals:
Answer: B. (2n)Cn
Why: Vandermonde identity: Σ(r=0 to n)(nCr)²=coeff of xⁿ in (1+x)<sup>n</sup>·(1+x)<sup>n</sup>=(1+x)<sup>2n</sup>, which equals (2n)Cn. Since nCr=nC(n−r), this is equivalent to coeff of xⁿ in (1+x)<sup>2n</sup>. Answer: (2n)Cn.
Q49.
In the expansion of (1 + x)ⁿ, if the coefficients of rth and (r+2)th terms are equal, then n =
Answer: C. 2r
Why: The rth term has coefficient C(n, r−1) and the (r+2)th term has coefficient C(n, r+1). Setting them equal: C(n, r−1) = C(n, r+1). Using the symmetry C(n,k) = C(n,n−k), this means r−1 = n−(r+1) = n−r−1, so 2r = n.
Q50.
Find the value of C₀ + C₁/2 + C₂/3 + ... + Cₙ/(n+1) where Cᵣ = nCr.
Answer: C. (2ⁿ⁺¹-1)/(n+1)
Why: Integrate (1+x)<sup>n</sup> from 0 to 1: [(1+x)<sup>n+1</sup>/(n+1)] from 0 to 1 = (2<sup>n+1</sup>-1)/(n+1). LHS integral = C<sub>0</sub> + C<sub>1</sub>/2 + C<sub>2</sub>/3 + ... + Cn/(n+1). So answer = (2<sup>n+1</sup>-1)/(n+1).
Q51.
The largest term in the expansion of (3 + 5x)¹⁰ when x = 1/5 is:
Answer: C. T₃
Why: With x=1/5, 5x=1, so T(r+1)=C(10,r)*3<sup>10-r</sup>. Computing directly: T<sub>1</sub>=3<sup>10</sup>=59049, T<sub>2</sub>=10*3<sup>9</sup>=196830, T3=45*3<sup>8</sup>=295245, T<sub>4</sub>=120*3<sup>7</sup>=262440. Since T3 exceeds both T<sub>2</sub> and T<sub>4</sub>, T3 is the largest term.
Q52.
The value of (nC<sub>1</sub>/nC<sub>0</sub>) + 2(nC<sub>2</sub>/nC<sub>1</sub>) + 3(nC<sub>3</sub>/nC<sub>2</sub>) + ... + n(nCn/nCn-1) is:
Answer: A. n(n+1)/2
Why: Simplify each ratio: nCr/nC(r−1)=(n−r+1)/r. So the r-th term is r·(n−r+1)/r=n−r+1. Sum=Σ(r=1 to n)(n−r+1)=n+(n−1)+…+1=n(n+1)/2. Uses consecutive-ratio reduction technique.
Q53.
In the expansion of (1 + x)<sup>n + 2</sup>, the coefficient of x<sup>n</sup> is (n+2)C n. Using this, find the coefficient of x² in (1 + x)⁵ + (1 + x)⁶ + ... + (1 + x)¹⁰.
Answer: B. 155
Why: Coefficient of x² in (1+x)<sup>k</sup> is kC2. Sum for k=5 to 10: 5C2+6C2+7C2+8C2+9C2+10C2 = 10+15+21+28+36+45 = 155. (By the hockey-stick identity, this also equals 11C3 - 5C3 = 165-10 = 155.)
Q54.
Using binomial theorem, the integral part of (5 + 2√6)ⁿ + (5 - 2√6)ⁿ is:
Answer: A. Even
Why: (5+2sqrt6)<sup>n</sup> + (5-2sqrt6)<sup>n</sup> is always an even integer because only even-power terms survive (irrational terms cancel), and the result is a sum of integers.
Q55.
If the middle term in (x + 1/x)<sup>2n</sup> is 184756 nCn, find n.
Answer: B. 10
Why: Even power 2n has (2n+1) terms; unique middle term is T(n+1)=C(2n,n)·xⁿ·(1/x)<sup>n</sup>=C(2n,n). Set C(2n,n)=184756. Trial n=10: C(20,10)=184756 ✓ (known value). Answer: n=10.
Q56.
The coefficient of x⁵ in (1 + x²)⁵(1 + x)⁴ is:
Answer: A. 60
Why: Coeff of x<sup>5</sup> in (1+x<sup>2</sup>)<sup>5</sup>(1+x)<sup>4</sup>: from (1+x<sup>2</sup>)<sup>5</sup>, only even powers x<sup>0</sup>, x<sup>2</sup>, x<sup>4</sup> are available (max x<sup>10</sup>), and from (1+x)<sup>4</sup> only powers up to x<sup>4</sup> are available. Pairing to total x<sup>5</sup>: x<sup>2</sup> from the first with x<sup>3</sup> from the second gives C(5,1)*C(4,3) = 5*4=20; x<sup>4</sup> from the first with x<sup>1</sup> from the second gives C(5,2)*C(4,1) = 10*4=40. Total = 20+40 = 60.
Q57.
If (1 + ax)ⁿ = 1 + 6x + 12x² + ..., find a and n.
Answer: A. a=2, n=3
Why: Coefficient of x: na = 6. Coefficient of x²: n(n-1)a²/2 = 12. From na=6, a=6/n. Substituting: n(n-1)(36/n²)/2=12, so 18(n-1)/n=12, giving 18n-18=12n, 6n=18, n=3, a=2.
Q58.
The sum of the series 1 + nC<sub>1</sub> + (nC<sub>2</sub>)² + ... + (nCn)² is:
Answer: A. (2n)Cn
Why: Recognise: 1=nC<sub>0</sub>=nC<sub>0</sub>², nC<sub>1</sub>=nC<sub>1</sub>² only if nC<sub>1</sub>=1 (n=1). In general the series is nC<sub>0</sub>²+nC<sub>1</sub>²+…+nCn². By Vandermonde identity this equals coeff of xⁿ in (1+x)<sup>2n</sup>=(2n)Cn. Answer: (2n)Cn.
Q59.
Find the numerically greatest term in (2 - 3x)¹⁰ when x = 1.
Answer: C. T₇
Why: |T(r+1)/Tr| = |(10-r+1)/r x (-3x/2)| = (11-r)/r x 3/2. Set >= 1: 3(11-r) >= 2r. 33-3r >= 2r. 33 >= 5r. r <= 6.6. Check r=6: T<sub>7</sub>/T<sub>6</sub> = 5/6 x 3/2 = 15/12 > 1, so T<sub>7</sub> > T<sub>6</sub>. Check r=7: T<sub>8</sub>/T<sub>7</sub> = 4/7 x 3/2 = 12/14 < 1, so T<sub>8</sub> < T<sub>7</sub>. So T<sub>7</sub> is the greatest term.
Q60.
The value of C₀ - C₁ + C₂ - ... + (−1)ⁿCₙ, where Cᵣ = nCr, is:
Answer: B. 0
Why: Substitute x=−1 into the binomial theorem (1+x)<sup>n</sup>=Σ(r=0 to n)nCr·xʳ: LHS=(1−1)<sup>n</sup>=0 (n≥1). RHS=C<sub>0</sub>−C<sub>1</sub>+C<sub>2</sub>−…+(−1)<sup>n</sup> Cn. Therefore the alternating sum =0 for all n≥1.
Q61.
The coefficient of x⁵ in the expansion of (1 + x)¹⁰ is:
Answer: B. 252
Why: C(10, 5) = 252.
Q62.
The term independent of x in the expansion of (x² − 1/x)⁶ is:
Answer: A. 15
Why: General term power is 12 − 3r; independent when r = 4, giving C(6,4)(−1)⁴ = 15.
Q63.
The middle term in the expansion of (2x − 1/x)⁸ is:
Answer: A. 1120
Why: With n = 8 the middle term is the 5th (r = 4): C(8,4)(2)⁴(−1)⁴ = 70·16 = 1120.
Q64.
The sum of all the binomial coefficients in the expansion of (2x + 3y)¹⁰ is:
Answer: A. 5¹⁰
Why: Setting x = y = 1 gives (2 + 3)¹⁰ = 5¹⁰.
Q65.
In the expansion of (1 + x)¹⁵, the coefficients of which two consecutive terms are equal?
Answer: B. x⁷ and x⁸
Why: C(15, r) = C(15, r+1) requires r + (r+1) = 15, so r = 7; the terms are x⁷ and x⁸.
Q66.
The value of C(10,0) + C(10,2) + C(10,4) + ... + C(10,10) is:
Answer: B. 512
Why: The sum of even-indexed binomial coefficients equals 2<sup>n−1</sup> = 2⁹ = 512.
Q67.
The last digit of 7¹⁰⁰ is:
Answer: A. 1
Why: Powers of 7 cycle with last digits 7, 9, 3, 1 (period 4). Since 100 is a multiple of 4, the last digit is 1.
Q68.
If (1 + x)ⁿ = C₀ + C₁x + ... + Cₙxⁿ, then C₀ + 2C₁ + 3C₂ + ... + (n+1)Cₙ equals:
Answer: A. 2<sup>n−1</sup>(n + 2)
Why: Σ(r+1)C(n,r) = n·2<sup>n−1</sup> + 2ⁿ = 2<sup>n−1</sup>(n + 2).