NCERT Solutions · Class 11 Mathematics Binomial Theorem

Difference of two expansions. Expand both by 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, the even-power terms \(\displaystyle a^4 \), \(\displaystyle 6a^2b^2 \), \(\displaystyle b^4 \) cancel and the odd ones double: \[(a+b)^4-(a-b)^4=8a^3b+8ab^3=8ab\left(a^2+b^2\right). \]Now put \(\displaystyle a=\sqrt{3} \), \(\displaystyle b=\sqrt{2} \), so \(\displaystyle ab=\sqrt{6} \) and \(\displaystyle a^2+b^2=3+2=5 \): \[(\sqrt{3}+\sqrt{2})^4-(\sqrt{3}-\sqrt{2})^4=8\sqrt{6}\times 5=40\sqrt{6}. \]As a check, \(\displaystyle 40\sqrt{6}\approx 40(2.4495)=97.98 \), which is what the two fourth powers differ by.\(\displaystyle (a+b)^4-(a-b)^4=8ab\left(a^2+b^2\right) \), and \(\displaystyle (\sqrt{3}+\sqrt{2})^4-(\sqrt{3}-\sqrt{2})^4=40\sqrt{6} \).

Sum of two expansions. Expand both by 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, the odd-power terms cancel and the even ones double: \[(x+1)^6+(x-1)^6=2\left(x^6+15x^4+15x^2+1\right). \]Now put \(\displaystyle x=\sqrt{2} \), so \(\displaystyle x^2=2 \), \(\displaystyle x^4=4 \), \(\displaystyle x^6=8 \): \[(\sqrt{2}+1)^6+(\sqrt{2}-1)^6=2\left(8+15(4)+15(2)+1\right)=2(8+60+30+1)=2(99)=198. \]\(\displaystyle (x+1)^6+(x-1)^6=2\left(x^6+15x^4+15x^2+1\right) \), and \(\displaystyle (\sqrt{2}+1)^6+(\sqrt{2}-1)^6=198 \).

Expand about a multiple of 8. Since \(\displaystyle 64=8^2 \), write \(\displaystyle 9 = 1+8 \) so that powers of \(\displaystyle 8 \) appear: \[9^{\,n+1}=(1+8)^{\,n+1}=\sum_{r=0}^{n+1}{}^{n+1}C_r\,8^{r} \] \[=1+{}^{n+1}C_1\,8+{}^{n+1}C_2\,8^2+{}^{n+1}C_3\,8^3+\cdots \] \[=1+8(n+1)+\sum_{r=2}^{n+1}{}^{n+1}C_r\,8^{r}. \]Therefore \[9^{\,n+1}-8n-9=\bigl(1+8n+8\bigr)-8n-9+\sum_{r=2}^{n+1}{}^{n+1}C_r\,8^{r}=\sum_{r=2}^{n+1}{}^{n+1}C_r\,8^{r}. \]Every remaining term has \(\displaystyle r\ge 2 \), so every term carries a factor \(\displaystyle 8^2=64 \): \[\sum_{r=2}^{n+1}{}^{n+1}C_r\,8^{r}=64\sum_{r=2}^{n+1}{}^{n+1}C_r\,8^{\,r-2}, \] and the sum on the right is an integer, since each \(\displaystyle {}^{n+1}C_r \) is an integer.(Check with \(\displaystyle n=1 \): \(\displaystyle 9^2-8-9=81-17=64 \), which is indeed \(\displaystyle 64\times 1 \).)Hence \(\displaystyle 9^{\,n+1}-8n-9 \) is a multiple of \(\displaystyle 64 \) for every positive integer \(\displaystyle n \).

Recognise the sum as a binomial expansion. The binomial theorem states \[(a+b)^n=\sum_{r=0}^{n}{}^{n}C_r\,a^{\,n-r}b^{\,r}. \]Choose \(\displaystyle a=1 \) and \(\displaystyle b=3 \). Then \(\displaystyle a^{\,n-r}=1 \) for every \(\displaystyle r \), and \(\displaystyle b^{r}=3^{r} \), so \[(1+3)^n=\sum_{r=0}^{n}{}^{n}C_r\,(1)^{\,n-r}3^{\,r}=\sum_{r=0}^{n}3^{r}\,{}^{n}C_r. \]The left side is \(\displaystyle (1+3)^n=4^n \).Hence \(\displaystyle \displaystyle\sum_{r=0}^{n}3^{r}\,{}^{n}C_r=4^{n} \).

Expand \(\displaystyle a \) about \(\displaystyle a-b \). Following the hint, write \(\displaystyle a=(a-b)+b \) and expand by the binomial theorem, treating \(\displaystyle a-b \) as the first term: \[a^{n}=\bigl((a-b)+b\bigr)^{n}=\sum_{r=0}^{n}{}^{n}C_r\,(a-b)^{r}\,b^{\,n-r}. \]Separate the \(\displaystyle r=0 \) term, which is \(\displaystyle {}^{n}C_0\,(a-b)^{0}b^{\,n}=b^{n} \): \[a^{n}=b^{n}+\sum_{r=1}^{n}{}^{n}C_r\,(a-b)^{r}\,b^{\,n-r}. \]Hence \[a^{n}-b^{n}=\sum_{r=1}^{n}{}^{n}C_r\,(a-b)^{r}\,b^{\,n-r}. \]Every term on the right has \(\displaystyle r\ge 1 \), so \(\displaystyle (a-b) \) can be taken out as a common factor: \[a^{n}-b^{n}=(a-b)\sum_{r=1}^{n}{}^{n}C_r\,(a-b)^{r-1}\,b^{\,n-r}=(a-b)\,k, \] where \(\displaystyle k=\sum_{r=1}^{n}{}^{n}C_r\,(a-b)^{r-1}b^{\,n-r} \) is an integer, because \(\displaystyle a \), \(\displaystyle b \) and every \(\displaystyle {}^{n}C_r \) are integers.So \(\displaystyle a-b \) is a factor of \(\displaystyle a^{n}-b^{n} \) for every positive integer \(\displaystyle n \).

Difference of two expansions. For a general pair \(\displaystyle a,b \), expanding to the sixth power and subtracting kills the even-power terms and doubles the odd ones: \[(a+b)^6-(a-b)^6=2\left[{}^{6}C_1a^5b+{}^{6}C_3a^3b^3+{}^{6}C_5ab^5\right] \] \[=2\left[6a^5b+20a^3b^3+6ab^5\right]. \]Now put \(\displaystyle a=\sqrt{3} \), \(\displaystyle b=\sqrt{2} \), and use \(\displaystyle a^5=9\sqrt{3} \), \(\displaystyle a^3=3\sqrt{3} \), \(\displaystyle b^3=2\sqrt{2} \), \(\displaystyle b^5=4\sqrt{2} \), \(\displaystyle \sqrt{3}\sqrt{2}=\sqrt{6} \): Their sum is \(\displaystyle (54+120+24)\sqrt{6}=198\sqrt{6} \), and doubling gives \(\displaystyle 396\sqrt{6} \).As a check, \(\displaystyle 396\sqrt{6}\approx 396(2.4495)\approx 970.0 \).\(\displaystyle (\sqrt{3}+\sqrt{2})^6-(\sqrt{3}-\sqrt{2})^6 = 396\sqrt{6} \)

Sum of two expansions. For a general pair \(\displaystyle A,B \), expanding to the fourth power and adding kills the odd-power terms and doubles the even ones: \[(A+B)^4+(A-B)^4=2\left[A^4+{}^{4}C_2A^2B^2+B^4\right]=2\left[A^4+6A^2B^2+B^4\right]. \]Put \(\displaystyle A=a^2 \) and \(\displaystyle B=\sqrt{a^2-1} \), so that \(\displaystyle B^2=a^2-1 \) and \(\displaystyle B^4=(a^2-1)^2 \): \[\left(a^{2}+\sqrt{a^{2}-1}\right)^{4}+\left(a^{2}-\sqrt{a^{2}-1}\right)^{4}=2\left[a^{8}+6a^{4}\left(a^{2}-1\right)+\left(a^{2}-1\right)^{2}\right]. \]Expand inside the bracket: \[a^{8}+6a^{6}-6a^{4}+a^{4}-2a^{2}+1=a^{8}+6a^{6}-5a^{4}-2a^{2}+1. \]Hence the value is \(\displaystyle 2a^{8}+12a^{6}-10a^{4}-4a^{2}+2 \).As a check, at \(\displaystyle a=2 \) the formula gives \(\displaystyle 512+768-160-16+2=1106 \), and direct computation of \(\displaystyle (4+\sqrt{3})^4+(4-\sqrt{3})^4 \) also gives \(\displaystyle 1106 \).\(\displaystyle \left(a^{2}+\sqrt{a^{2}-1}\right)^{4}+\left(a^{2}-\sqrt{a^{2}-1}\right)^{4}=2a^{8}+12a^{6}-10a^{4}-4a^{2}+2 \)

Truncated binomial expansion. Write \(\displaystyle 0.99 = 1-0.01 \), so the terms shrink quickly: \[(0.99)^5=(1-0.01)^5={}^{5}C_0-{}^{5}C_1(0.01)+{}^{5}C_2(0.01)^2-\cdots \]Keeping the first three terms: \[(0.99)^5\approx 1-5(0.01)+10(0.0001)=1-0.05+0.001=0.951. \]The next term is \(\displaystyle -{}^{5}C_3(0.01)^3=-10(0.000001)=-0.00001 \), which is why three terms already fix the answer to three decimal places. (The exact value is \(\displaystyle 0.9509900499 \).)\(\displaystyle (0.99)^5 \approx 0.951 \)

Binomial Theorem, by grouping. The binomial theorem expands a two-term power, so first turn the trinomial into a binomial by keeping \(\displaystyle 1+\frac{x}{2}\) together as one term:\[\left(1+\frac{x}{2}-\frac{2}{x}\right)^{4}=\left[\left(1+\frac{x}{2}\right)-\frac{2}{x}\right]^{4}\]Apply \(\displaystyle (p-q)^{4}=p^{4}-4p^{3}q+6p^{2}q^{2}-4pq^{3}+q^{4}\) with \(\displaystyle p=1+\frac{x}{2}\) and \(\displaystyle q=\frac{2}{x}\):\[=\left(1+\frac{x}{2}\right)^{4}-\frac{8}{x}\left(1+\frac{x}{2}\right)^{3}+\frac{24}{x^{2}}\left(1+\frac{x}{2}\right)^{2}-\frac{32}{x^{3}}\left(1+\frac{x}{2}\right)+\frac{16}{x^{4}}\]Now expand each power of \(\displaystyle 1+\frac{x}{2}\), again by the binomial theorem:\[\left(1+\frac{x}{2}\right)^{4}=1+2x+\frac{3x^{2}}{2}+\frac{x^{3}}{2}+\frac{x^{4}}{16},\qquad \left(1+\frac{x}{2}\right)^{3}=1+\frac{3x}{2}+\frac{3x^{2}}{4}+\frac{x^{3}}{8}\]\[\left(1+\frac{x}{2}\right)^{2}=1+x+\frac{x^{2}}{4}\]Substituting and clearing each bracket:\[-\frac{8}{x}\left(1+\frac{3x}{2}+\frac{3x^{2}}{4}+\frac{x^{3}}{8}\right)=-\frac{8}{x}-12-6x-x^{2}\]\[\frac{24}{x^{2}}\left(1+x+\frac{x^{2}}{4}\right)=\frac{24}{x^{2}}+\frac{24}{x}+6,\qquad -\frac{32}{x^{3}}\left(1+\frac{x}{2}\right)=-\frac{32}{x^{3}}-\frac{16}{x^{2}}\]Collect like powers of \(\displaystyle x\): \(\displaystyle \left(1+\dfrac{x}{2}-\dfrac{2}{x}\right)^{4}=\dfrac{x^{4}}{16}+\dfrac{x^{3}}{2}+\dfrac{x^{2}}{2}-4x-5+\dfrac{16}{x}+\dfrac{8}{x^{2}}-\dfrac{32}{x^{3}}+\dfrac{16}{x^{4}}\)(Check at \(\displaystyle x=2\): the bracket is \(\displaystyle 1+1-1=1\), so the value must be \(\displaystyle 1\); the expansion gives \(\displaystyle 1+4+2-8-5+8+2-4+1=1\).)

Binomial Theorem, by grouping. Make the trinomial a binomial by pulling \(\displaystyle a\) out of the two terms that carry it:\[\left(3x^{2}-2ax+3a^{2}\right)^{3}=\left[3x^{2}-a(2x-3a)\right]^{3}\]Apply \(\displaystyle (p-q)^{3}=p^{3}-3p^{2}q+3pq^{2}-q^{3}\) with \(\displaystyle p=3x^{2}\) and \(\displaystyle q=a(2x-3a)\):\[=27x^{6}-3\left(9x^{4}\right)a(2x-3a)+3\left(3x^{2}\right)a^{2}(2x-3a)^{2}-a^{3}(2x-3a)^{3}\]Expand the powers of \(\displaystyle 2x-3a\), once more by the binomial theorem:\[(2x-3a)^{2}=4x^{2}-12ax+9a^{2},\qquad (2x-3a)^{3}=8x^{3}-36ax^{2}+54a^{2}x-27a^{3}\]Substituting each piece:\[-27x^{4}\,a(2x-3a)=-54ax^{5}+81a^{2}x^{4}\]\[9a^{2}x^{2}\left(4x^{2}-12ax+9a^{2}\right)=36a^{2}x^{4}-108a^{3}x^{3}+81a^{4}x^{2}\]\[-a^{3}\left(8x^{3}-36ax^{2}+54a^{2}x-27a^{3}\right)=-8a^{3}x^{3}+36a^{4}x^{2}-54a^{5}x+27a^{6}\]Collect like terms: \(\displaystyle \left(3x^{2}-2ax+3a^{2}\right)^{3}=27x^{6}-54ax^{5}+117a^{2}x^{4}-116a^{3}x^{3}+117a^{4}x^{2}-54a^{5}x+27a^{6}\)(The coefficients read the same forwards and backwards, as they must: swapping \(\displaystyle x\leftrightarrow a\) leaves \(\displaystyle 3x^{2}-2ax+3a^{2}\) unchanged.)