Expand (1-2x)^5.
Hint. Use row 5 of Pascal's triangle (1,5,10,10,5,1) with b=-2x, so the signs alternate.
With a=1, b=-2x, n=5: C0(1)^5 + C1(1)^4(-2x) + C2(1)^3(-2x)^2 + C3(1)^2(-2x)^3 + C4(1)(-2x)^4 + C5(-2x)^5 = 1 -5(2x) +10(4x^2) -10(8x^3) +5(16x^4) -32x^5 = 1-10x+40x^2-80x^3+80x^4-32x^5.
✦ Working through each part gives: 1-10x+40x^2-80x^3+80x^4-32x^5.
Expand (2/x - x/2)^5.
Hint. With a=2/x and b=-x/2, apply row 5 of Pascal's triangle and simplify each power of a and b separately before multiplying.
With a=2/x, b=-x/2, n=5: T0=C0.a^5=(2/x)^5=32/x^5. T1=C1.a^4.b=5(16/x^4)(-x/2)=-40/x^3. T2=C2.a^3.b^2=10(8/x^3)(x^2/4)=20/x. T3=C3.a^2.b^3=10(4/x^2)(-x^3/8)=-5x. T4=C4.a.b^4=5(2/x)(x^4/16)=5x^3/8. T5=C5.b^5=(-x/2)^5=-x^5/32. Adding all six terms gives the expansion.
✦ Working through each part gives: 32/x^5 - 40/x^3 + 20/x - 5x + 5x^3/8 - x^5/32.
Expand (2x-3)^6.
Hint. Use row 6 of Pascal's triangle (1,6,15,20,15,6,1) with b=-3, so signs alternate starting positive.
With a=2x, b=-3, n=6: (2x)^6 + 6(2x)^5(-3) + 15(2x)^4(9) + 20(2x)^3(-27) + 15(2x)^2(81) + 6(2x)(-243) + 729 = 64x^6-576x^5+2160x^4-4320x^3+4860x^2-2916x+729.
✦ Working through each part gives: 64x^6-576x^5+2160x^4-4320x^3+4860x^2-2916x+729.
Expand (x/3 + 1/x)^5.
Hint. Use row 5 of Pascal's triangle with a=x/3 and b=1/x; every term is positive since both a and b are positive.
With a=x/3, b=1/x, n=5: (x/3)^5 + 5(x/3)^4(1/x) + 10(x/3)^3(1/x)^2 + 10(x/3)^2(1/x)^3 + 5(x/3)(1/x)^4 + (1/x)^5 = x^5/243 + 5x^3/81 + 10x/27 + 10/(9x) + 5/(3x^3) + 1/x^5.
✦ Working through each part gives: x^5/243 + 5x^3/81 + 10x/27 + 10/(9x) + 5/(3x^3) + 1/x^5.
Expand (x + 1/x)^6.
Hint. Use row 6 of Pascal's triangle with a=x and b=1/x; every term is positive.
With a=x, b=1/x, n=6: x^6 + 6x^5(1/x) + 15x^4(1/x)^2 + 20x^3(1/x)^3 + 15x^2(1/x)^4 + 6x(1/x)^5 + (1/x)^6 = x^6+6x^4+15x^2+20+15/x^2+6/x^4+1/x^6.
✦ Working through each part gives: x^6+6x^4+15x^2+20+15/x^2+6/x^4+1/x^6.
Using the Binomial Theorem, evaluate (96)^3.
Hint. Write 96 as 100-4, so only four small terms need to be computed exactly.
(100-4)^3 = C0(100)^3 - C1(100)^2(4) + C2(100)(4)^2 - C3(4)^3 = 1000000 - 120000 + 4800 - 64 = 884736.
✦ Working through each part gives: 884736.
Using the Binomial Theorem, evaluate (102)^5.
Hint. Write 102 as 100+2, so every term is positive and only six terms need to be summed.
(100+2)^5 = C0(100)^5 + C1(100)^4(2) + C2(100)^3(4) + C3(100)^2(8) + C4(100)(16) + C5(32) = 10000000000+1000000000+40000000+800000+8000+32 = 11040808032.
✦ Working through each part gives: 11040808032.
Using the Binomial Theorem, evaluate (101)^4.
Hint. Write 101 as 100+1, which keeps every term simple to compute.
(100+1)^4 = C0(100)^4 + C1(100)^3 + C2(100)^2 + C3(100) + C4 = 100000000+4000000+60000+400+1 = 104060401.
✦ Working through each part gives: 104060401.
Using the Binomial Theorem, evaluate (99)^5.
Hint. Write 99 as 100-1, so the signs alternate but the magnitudes stay simple.
(100-1)^5 = C0(100)^5 - C1(100)^4 + C2(100)^3 - C3(100)^2 + C4(100) - C5 = 10000000000-500000000+10000000-100000+500-1 = 9509900499.
✦ Working through each part gives: 9509900499.
Using Binomial Theorem, indicate which number is larger, (1.1)^10000 or 1000.
Hint. Write 1.1 as 1+0.1 and expand -- the first two terms alone already decide the comparison, since every remaining term is positive.
(1.1)^10000 = (1+0.1)^10000 = C0 + C1(0.1) + other positive terms = 1 + 10000(0.1) + other positive terms = 1+1000+other positive terms > 1000.
✦ Working through each part gives: (1.1)^10000 is larger than 1000.
Find (a+b)^4 - (a-b)^4. Hence, evaluate (sqrt3+sqrt2)^4 - (sqrt3-sqrt2)^4.
Hint. Expand both fourth powers fully and subtract -- the even-r terms cancel, leaving only the odd-r terms doubled.
(a+b)^4=a^4+4a^3b+6a^2b^2+4ab^3+b^4 and (a-b)^4=a^4-4a^3b+6a^2b^2-4ab^3+b^4. Subtracting cancels the even-r terms: (a+b)^4-(a-b)^4 = 8a^3b+8ab^3 = 8ab(a^2+b^2). At a=sqrt3, b=sqrt2: 8(sqrt3)(sqrt2)(3+2) = 8.sqrt6.5 = 40.sqrt6.
✦ Working through each part gives: (a+b)^4-(a-b)^4 = 8ab(a^2+b^2); (sqrt3+sqrt2)^4-(sqrt3-sqrt2)^4 = 40.sqrt6.
Find (x+1)^6 + (x-1)^6. Hence or otherwise evaluate (sqrt2+1)^6 + (sqrt2-1)^6.
Hint. Expand both sixth powers fully and add -- the odd-r terms cancel, leaving only the even-r terms doubled.
(x+1)^6=x^6+6x^5+15x^4+20x^3+15x^2+6x+1 and (x-1)^6=x^6-6x^5+15x^4-20x^3+15x^2-6x+1. Adding cancels the odd-r terms: (x+1)^6+(x-1)^6 = 2x^6+30x^4+30x^2+2 = 2(x^6+15x^4+15x^2+1). At x=sqrt2: x^2=2,x^4=4,x^6=8, giving 2(8+60+30+1)=2(99)=198.
✦ Working through each part gives: (x+1)^6+(x-1)^6 = 2(x^6+15x^4+15x^2+1); (sqrt2+1)^6+(sqrt2-1)^6 = 198.
Show that 9^(n+1) - 8n - 9 is divisible by 64, whenever n is a positive integer.
Hint. Write 9=1+8 and expand (1+8)^(n+1) with the Binomial Theorem; isolate the first two terms and show the rest carries a factor of 64.
9^(n+1) = (1+8)^(n+1) = C(n+1,0) + C(n+1,1)(8) + C(n+1,2)(8)^2 + ... + C(n+1,n+1)(8)^(n+1) = 1 + 8(n+1) + 64[C(n+1,2) + 8.C(n+1,3) + ... + 8^(n-1).C(n+1,n+1)] = 9 + 8n + 64K, where K is a natural number. So 9^(n+1) - 8n - 9 = 64K, a multiple of 64.
✦ Working through each part gives: proved: 9^(n+1)-8n-9 = 64K for a natural number K, so it is always divisible by 64.
Where students slip. Forgetting that the first two terms of (1+8)^(n+1) combine with the -8n-9 being subtracted to leave exactly 0 constant and 0 n-term -- the whole point is that only the factor-of-64 remainder survives, so both isolated terms must cancel exactly.
Prove that the sum from r=0 to n of 3^r.nCr = 4^n.
Hint. Recognise this sum as the right side of the Binomial Theorem for (1+3)^n with a=1, b=3.
By the Binomial Theorem, (1+3)^n = sum_(r=0)^n nCr.(1)^(n-r).(3)^r = sum_(r=0)^n 3^r.nCr. Since 1+3=4, the left side is also 4^n. So sum_(r=0)^n 3^r.nCr = 4^n.
✦ Working through each part gives: proved: sum_(r=0)^n 3^r.nCr = 4^n, directly from (1+3)^n=4^n.
Solutions written by the tuition.in editorial team and checked against the NCERT Class 11 Mathematics textbook, Reprint 2026-27 (kemh107.pdf) — a 9-page, single-exercise chapter (Exercise 7.1, 14 questions) plus a 6-question Miscellaneous Exercise, 20 questions total. Confirmed against the CBSE curriculum PDF that the general term and middle term of a binomial expansion are formative-only content for this chapter (not summative), matching the current book's complete absence of any such section or a second numbered exercise. Several stacked-fraction and trinomial questions (Ex 7.1 Q2, Q4, Q5; Misc Q3, Q5, Q6) were cross-checked via 300dpi page renders, and the two trinomial-expansion answers (Misc Q5, Q6) were independently verified by numeric substitution at multiple values before being finalised. Questions are referenced from the NCERT textbook for identification.
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