Where Mindful Learning Begins
Where Mindful Learning Begins
Free, step-by-step Class 11 Maths NCERT Solutions for Chapter 7 Ex 7.1 — all 14 questions solved, covering binomial expansions, computing large powers, and divisibility proofs using the Binomial Theorem.
Every question in this exercise comes back to one formula: (a+b)^n=\sum_{r=0}^{n}{}^nC_r\,a^{n-r}b^r. Questions 1–5 are direct expansions, best organised by writing out the row of Pascal's Triangle for the given index first. Questions 6–10 use the same idea in reverse — splitting an awkward number like 96 or 102 into a sum or difference (100 − 4, 100 + 2) that expands cleanly. Questions 11–14 go a step further, using the expansion algebraically to derive shortcuts (like cancelling every odd-power term) and to prove divisibility and summation identities.
Using the Binomial Theorem with a=1, b=-2x, n=5, and the row 1, 5, 10, 10, 5, 1 from Pascal's Triangle:
(1-2x)^5={}^5C_0(1)^5+{}^5C_1(1)^4(-2x)+{}^5C_2(1)^3(-2x)^2+{}^5C_3(1)^2(-2x)^3+{}^5C_4(1)(-2x)^4+{}^5C_5(-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
Using the Binomial Theorem with a=\dfrac{2}{x}, b=-\dfrac{x}{2}, n=5:
\left(\dfrac{2}{x}-\dfrac{x}{2}\right)^5={}^5C_0\left(\dfrac{2}{x}\right)^5+{}^5C_1\left(\dfrac{2}{x}\right)^4\left(-\dfrac{x}{2}\right)+{}^5C_2\left(\dfrac{2}{x}\right)^3\left(-\dfrac{x}{2}\right)^2+{}^5C_3\left(\dfrac{2}{x}\right)^2\left(-\dfrac{x}{2}\right)^3+{}^5C_4\left(\dfrac{2}{x}\right)\left(-\dfrac{x}{2}\right)^4+{}^5C_5\left(-\dfrac{x}{2}\right)^5
Simplifying each term:
=\dfrac{32}{x^5}+5\times\dfrac{16}{x^4}\times\left(-\dfrac{x}{2}\right)+10\times\dfrac{8}{x^3}\times\dfrac{x^2}{4}+10\times\dfrac{4}{x^2}\times\left(-\dfrac{x^3}{8}\right)+5\times\dfrac{2}{x}\times\dfrac{x^4}{16}+\left(-\dfrac{x^5}{32}\right)
=\dfrac{32}{x^5}-\dfrac{40}{x^3}+\dfrac{20}{x}-5x+\dfrac{5x^3}{8}-\dfrac{x^5}{32}
Using the Binomial Theorem with a=2x, b=-3, n=6, and the row 1, 6, 15, 20, 15, 6, 1:
(2x-3)^6={}^6C_0(2x)^6+{}^6C_1(2x)^5(-3)+{}^6C_2(2x)^4(-3)^2+{}^6C_3(2x)^3(-3)^3+{}^6C_4(2x)^2(-3)^4+{}^6C_5(2x)(-3)^5+{}^6C_6(-3)^6
=64x^6+6(32x^5)(-3)+15(16x^4)(9)+20(8x^3)(-27)+15(4x^2)(81)+6(2x)(-243)+729
=64x^6-576x^5+2160x^4-4320x^3+4860x^2-2916x+729
Using the Binomial Theorem with a=\dfrac{x}{3}, b=\dfrac{1}{x}, n=5:
\left(\dfrac{x}{3}+\dfrac{1}{x}\right)^5={}^5C_0\left(\dfrac{x}{3}\right)^5+{}^5C_1\left(\dfrac{x}{3}\right)^4\left(\dfrac{1}{x}\right)+{}^5C_2\left(\dfrac{x}{3}\right)^3\left(\dfrac{1}{x}\right)^2+{}^5C_3\left(\dfrac{x}{3}\right)^2\left(\dfrac{1}{x}\right)^3+{}^5C_4\left(\dfrac{x}{3}\right)\left(\dfrac{1}{x}\right)^4+{}^5C_5\left(\dfrac{1}{x}\right)^5
Simplifying each term:
=\dfrac{x^5}{243}+5\times\dfrac{x^4}{81}\times\dfrac{1}{x}+10\times\dfrac{x^3}{27}\times\dfrac{1}{x^2}+10\times\dfrac{x^2}{9}\times\dfrac{1}{x^3}+5\times\dfrac{x}{3}\times\dfrac{1}{x^4}+\dfrac{1}{x^5}
=\dfrac{x^5}{243}+\dfrac{5x^3}{81}+\dfrac{10x}{27}+\dfrac{10}{9x}+\dfrac{5}{3x^3}+\dfrac{1}{x^5}
Using the Binomial Theorem with a=x, b=\dfrac{1}{x}, n=6, and the row 1, 6, 15, 20, 15, 6, 1:
\left(x+\dfrac{1}{x}\right)^6={}^6C_0x^6+{}^6C_1x^5\left(\dfrac{1}{x}\right)+{}^6C_2x^4\left(\dfrac{1}{x}\right)^2+{}^6C_3x^3\left(\dfrac{1}{x}\right)^3+{}^6C_4x^2\left(\dfrac{1}{x}\right)^4+{}^6C_5x\left(\dfrac{1}{x}\right)^5+{}^6C_6\left(\dfrac{1}{x}\right)^6
=x^6+6x^4+15x^2+20+\dfrac{15}{x^2}+\dfrac{6}{x^4}+\dfrac{1}{x^6}
Write 96=100-4, so (96)^3=(100-4)^3.
(100-4)^3={}^3C_0(100)^3-{}^3C_1(100)^2(4)+{}^3C_2(100)(4)^2-{}^3C_3(4)^3
=1000000-3(10000)(4)+3(100)(16)-64
=1000000-120000+4800-64
Write 102=100+2, so (102)^5=(100+2)^5.
(100+2)^5={}^5C_0(100)^5+{}^5C_1(100)^4(2)+{}^5C_2(100)^3(2)^2+{}^5C_3(100)^2(2)^3+{}^5C_4(100)(2)^4+{}^5C_5(2)^5
=10000000000+5(100000000)(2)+10(1000000)(4)+10(10000)(8)+5(100)(16)+32
=10000000000+1000000000+40000000+800000+8000+32
Write 101=100+1, so (101)^4=(100+1)^4.
(100+1)^4={}^4C_0(100)^4+{}^4C_1(100)^3(1)+{}^4C_2(100)^2(1)^2+{}^4C_3(100)(1)^3+{}^4C_4(1)^4
=100000000+4(1000000)+6(10000)+4(100)+1
=100000000+4000000+60000+400+1
Write 99=100-1, so (99)^5=(100-1)^5.
(100-1)^5={}^5C_0(100)^5-{}^5C_1(100)^4(1)+{}^5C_2(100)^3(1)^2-{}^5C_3(100)^2(1)^3+{}^5C_4(100)(1)^4-{}^5C_5(1)^5
=10000000000-500000000+10000000-100000+500-1
Write 1.1=1+0.1, so (1.1)^{10000}=(1+0.1)^{10000}.
Expanding just the first two terms of the binomial expansion (every remaining term is positive):
(1+0.1)^{10000}={}^{10000}C_0+{}^{10000}C_1(0.1)+\text{other positive terms}
=1+10000\times0.1+\text{other positive terms}=1+1000+\text{other positive terms}
Since all the remaining terms in the expansion are positive, this sum is strictly greater than 1001, which is already greater than 1000.
Expanding both using the Binomial Theorem:
(a+b)^4=a^4+4a^3b+6a^2b^2+4ab^3+b^4
(a-b)^4=a^4-4a^3b+6a^2b^2-4ab^3+b^4
Subtracting, every even-power term (in b) cancels, leaving only the odd-power terms doubled:
(a+b)^4-(a-b)^4=8a^3b+8ab^3=8ab(a^2+b^2)
Now substitute a=\sqrt3, b=\sqrt2:
8ab(a^2+b^2)=8(\sqrt3)(\sqrt2)\left((\sqrt3)^2+(\sqrt2)^2\right)=8\sqrt6\,(3+2)=40\sqrt6
Expanding both using the Binomial Theorem:
(x+1)^6=x^6+6x^5+15x^4+20x^3+15x^2+6x+1
(x-1)^6=x^6-6x^5+15x^4-20x^3+15x^2-6x+1
Adding, every odd-power term cancels, leaving only the even-power terms doubled:
(x+1)^6+(x-1)^6=2x^6+30x^4+30x^2+2=2(x^6+15x^4+15x^2+1)
Now substitute x=\sqrt2, so x^2=2, x^4=4, x^6=8:
2(8+15(4)+15(2)+1)=2(8+60+30+1)=2(99)=198
To show 9^{n+1}-8n-9 is divisible by 64, it is enough to write it in the form 64k for some natural number k.
Write 9=1+8, so 9^n=(1+8)^n. By the Binomial Theorem:
(1+8)^n={}^nC_0+{}^nC_1(8)+{}^nC_2(8)^2+\ldots+{}^nC_n(8)^n
9^n=1+8n+8^2\left[{}^nC_2+{}^nC_3(8)+\ldots+{}^nC_n(8)^{n-2}\right]
Let M={}^nC_2+8\,{}^nC_3+\ldots+8^{n-2}\,{}^nC_n, so 9^n=1+8n+64M.
Multiplying both sides by 9:
9^{n+1}=9(1+8n+64M)=9+72n+576M
Therefore:
9^{n+1}-8n-9=9+72n+576M-8n-9=64n+576M=64(n+9M)
By the Binomial Theorem, with a=1 and b=3:
(1+3)^n=\sum_{r=0}^{n}{}^nC_r(1)^{n-r}(3)^r=\sum_{r=0}^{n}3^r\,{}^nC_r
Since 1+3=4, the left-hand side is simply 4^n.
Therefore:
\sum_{r=0}^{n}3^r\,{}^nC_r=4^n
Expert CBSE Coaching · Class 9–12