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 \(\displaystyle \displaystyle\sum_{r=0}^{n}3^{r}\,{}^{n}C_r=4^{n} \).