math_binomialBinomial Theorem
Pascal triangle$\begin{array}{ccccccccccccccc}0& & & & & & & & 1& & & & & & \\ 1& & & & & & & 1& & 1& & & & & \\ 2& & & & & & 1& & 2& & 1& & & & \\ 3& & & & & 1& & 3& & 3& & 1& & & \\ 4& & & & 1& & 4& & 6& & 4& & 1& & \\ 5& & & 1& & 5& & 10& & 10& & 5& & 1& \\ 6& & 1& & 6& & 15& & 20& & 15& & 6& & 1\end{array}$ $\left(a+b{\right)}^{6}={a}^{6}+6{a}^{5}b+15{a}^{4}{b}^{2}+20{a}^{3}{b}^{3}+15{a}^{2}{b}^{4}+6a{b}^{5}+{b}^{6}$
Binomial theorem of any positive integer n$\left(a+b{\right)}^{n}{=}^{n}{C}_{0}{a}^{n}{b}^{0}{+}^{n}{C}_{1}{a}^{n-1}{b}^{1}{+}^{n}{C}_{2}{a}^{n-2}{b}^{2}+\phantom{\rule{0.22em}{0ex}}.\phantom{\rule{0.22em}{0ex}}.+{\phantom{\rule{0.22em}{0ex}}}^{n}{C}_{n-1}{a}^{1}{b}^{n-1}+{\phantom{\rule{0.22em}{0ex}}}^{n}{C}_{n}{a}^{0}{b}^{n}$
Binomial theorem of any positive integer n$\left(a+b{\right)}^{n}=\underset{k=0}{\overset{n}{âˆ‘}}{\phantom{\rule{0.22em}{0ex}}}^{n}{C}_{k}{a}^{n-k}{b}^{k}\phantom{\rule{0.22em}{0ex}}$
General Term${T}_{r+1}{=}^{n}{C}_{r}{a}^{n-r}{b}^{r}\phantom{\rule{0.22em}{0ex}}$
Middle term of $\left(a+b{\right)}^{2}$ when n is oddThere will be two middle term $\left(\frac{n+1}{2}{\right)}^{th}\phantom{\rule{0.22em}{0ex}}term\phantom{\rule{0.44em}{0ex}}and\phantom{\rule{0.22em}{0ex}}\left(\frac{n+1}{2}+1{\right)}^{th}\phantom{\rule{0.22em}{0ex}}term\phantom{\rule{0.44em}{0ex}}$
Middle term of $\left(a+b{\right)}^{2}$ when n is evenThere will be two middle term $\phantom{\rule{0.22em}{0ex}}\left(\frac{n}{2}+1{\right)}^{th}\phantom{\rule{0.22em}{0ex}}term\phantom{\rule{0.44em}{0ex}}$
Q.ex.4 (NCERT)Using binomial theorem, prove that ${6}^{n}-5n$ always leaves remainder 1 when divided by 25.A.ex.4${6}^{n}-5n=\left(1+5{\right)}^{n}-5n\phantom{\rule{0.44em}{0ex}}$$âŸ¹{6}^{n}-5n{=}^{n}{C}_{0}{+}^{n}{C}_{1}{5}^{1}{+}^{n}{C}_{2}{5}^{2}+\phantom{\rule{0.22em}{0ex}}.\phantom{\rule{0.22em}{0ex}}.+{\phantom{\rule{0.22em}{0ex}}}^{n}{C}_{n-1}{5}^{n-1}+{\phantom{\rule{0.22em}{0ex}}}^{n}{C}_{n}{5}^{n}-5n$$âŸ¹{6}^{n}-5n=1+5n{+}^{n}{C}_{2}{5}^{2}+\phantom{\rule{0.22em}{0ex}}.\phantom{\rule{0.22em}{0ex}}.+{\phantom{\rule{0.22em}{0ex}}}^{n}{C}_{n-1}{5}^{n-1}+{\phantom{\rule{0.22em}{0ex}}}^{n}{C}_{n}{5}^{n}-5n$$âŸ¹{6}^{n}-5n=1+{5}^{2}{\left(}^{n}{C}_{2}+\phantom{\rule{0.22em}{0ex}}.\phantom{\rule{0.22em}{0ex}}.+{\phantom{\rule{0.22em}{0ex}}}^{n}{C}_{n-1}{5}^{n-3}+{\phantom{\rule{0.22em}{0ex}}}^{n}{C}_{n}{5}^{n-2}\right)\phantom{\rule{0.22em}{0ex}}$$âŸ¹{6}^{n}-5n=1+25{\left(}^{n}{C}_{2}+\phantom{\rule{0.22em}{0ex}}.\phantom{\rule{0.22em}{0ex}}.+{\phantom{\rule{0.22em}{0ex}}}^{n}{C}_{n-1}{5}^{n-3}+{\phantom{\rule{0.22em}{0ex}}}^{n}{C}_{n}{5}^{n-2}\right)\phantom{\rule{0.22em}{0ex}}$$âŸ¹{6}^{n}-5n\phantom{\rule{0.22em}{0ex}}is\phantom{\rule{0.22em}{0ex}}in\phantom{\rule{0.22em}{0ex}}form\phantom{\rule{0.22em}{0ex}}1+25k\phantom{\rule{0.22em}{0ex}}$$âŸ¹\phantom{\rule{0.22em}{0ex}}so,\phantom{\rule{0.22em}{0ex}}{6}^{n}-5n\phantom{\rule{0.44em}{0ex}}is\phantom{\rule{0.22em}{0ex}}divisible\phantom{\rule{0.22em}{0ex}}by\phantom{\rule{0.22em}{0ex}}25\phantom{\rule{0.44em}{0ex}}$ Q.8.misc.4If a and b are distinct integers, prove that (a â€“ b) is a factor of ${a}^{n}\phantom{\rule{0.22em}{0ex}}\mathrm{â€“}\phantom{\rule{0.22em}{0ex}}{b}^{n}$, whenevern is a positive integer.A.8.misc.4${a}^{n}\phantom{\rule{0.22em}{0ex}}\mathrm{â€“}\phantom{\rule{0.22em}{0ex}}{b}^{n}=\left(a-b+b{\right)}^{n}-{b}^{n}$$âŸ¹{a}^{n}\phantom{\rule{0.22em}{0ex}}\mathrm{â€“}\phantom{\rule{0.22em}{0ex}}{b}^{n}{=}^{n}{C}_{0}\left(a-b{\right)}^{n}{b}^{0}{+}^{n}{C}_{1}\left(a-b{\right)}^{n-1}{b}^{1}+\phantom{\rule{0.22em}{0ex}}.\phantom{\rule{0.22em}{0ex}}.+{\phantom{\rule{0.22em}{0ex}}}^{n}{C}_{n-1}\left(a-b{\right)}^{1}{b}^{n-1}+{\phantom{\rule{0.22em}{0ex}}}^{n}{C}_{n}\left(a-b{\right)}^{0}{b}^{n}-{b}^{n}$$âŸ¹{a}^{n}\phantom{\rule{0.22em}{0ex}}\mathrm{â€“}\phantom{\rule{0.22em}{0ex}}{b}^{n}{=}^{n}{C}_{0}\left(a-b{\right)}^{n}{b}^{0}{+}^{n}{C}_{1}\left(a-b{\right)}^{n-1}{b}^{1}+\phantom{\rule{0.22em}{0ex}}.\phantom{\rule{0.22em}{0ex}}.+{\phantom{\rule{0.22em}{0ex}}}^{n}{C}_{n-1}\left(a-b{\right)}^{1}{b}^{n-1}+\phantom{\rule{0.22em}{0ex}}{b}^{n}-{b}^{n}$$âŸ¹{a}^{n}\phantom{\rule{0.22em}{0ex}}\mathrm{â€“}\phantom{\rule{0.22em}{0ex}}{b}^{n}=\left(a-b\right){\left\{}^{n}{C}_{0}\left(a-b{\right)}^{n-1}{b}^{0}{+}^{n}{C}_{1}\left(a-b{\right)}^{n-2}{b}^{1}+\phantom{\rule{0.22em}{0ex}}.\phantom{\rule{0.22em}{0ex}}.+{\phantom{\rule{0.22em}{0ex}}}^{n}{C}_{n-1}{b}^{n-1}\right\}$So, (a-b) is factor of ${a}^{n}\phantom{\rule{0.22em}{0ex}}\mathrm{â€“}\phantom{\rule{0.22em}{0ex}}{b}^{n}$So, ${a}^{n}\phantom{\rule{0.22em}{0ex}}\mathrm{â€“}\phantom{\rule{0.22em}{0ex}}{b}^{n}$ is divisible by (a-b) Q.8.misc.5Evaluate $\left(\sqrt{3}+\sqrt{2}{\right)}^{6}-\left(\sqrt{3}-\sqrt{2}{\right)}^{6}$A.8.misc.5$\left(\sqrt{3}+\sqrt{2}{\right)}^{6}-\left(\sqrt{3}-\sqrt{2}{\right)}^{6}$$=\left\{\left(\sqrt{3}+\sqrt{2}{\right)}^{2}{\right\}}^{3}-\left\{\left(\sqrt{3}-\sqrt{2}{\right)}^{2}{\right\}}^{3}$$=\left(5+2\sqrt{6}{\right)}^{3}-\left(5-2\sqrt{6}{\right)}^{3}$$=\left\{\left(5+2\sqrt{6}\right)-\left(5-2\sqrt{6}\right)\right\}\left\{\phantom{\rule{0.22em}{0ex}}\left(5+2\sqrt{6}{\right)}^{2}+\left(5+2\sqrt{6}\right)\left(5-2\sqrt{6}\right)+\left(5-2\sqrt{6}{\right)}^{2}\phantom{\rule{0.22em}{0ex}}\right\}$$=\left(4\sqrt{6}\right)\left\{25+24+20\sqrt{6}+25-24+25+24-20\sqrt{6}\right\}$$=\left(4\sqrt{6}\right)\left(99\right)=396\sqrt{6}$