# Binomial Theorem Questions

Solution : The given expression is $$(3x – {x^3\over 6})^7$$. Here n = 7, which is an odd number. By using middle terms in binomial expansion formula, So, $$({7 + 1\over 2}$$) th and $$({7 + 1\over 2} + 1)$$ th i.e.  4th and 5th terms are two middle terms. Now, $$T_{4}$$ = $$T_{3 + … ## Find the middle term in the expansion of \(({2\over 3}x^2 – {3\over 2x})^{20}$$.

Solution : Here, n = 20, which is an even number. So, $${20\over 2} + 1$$th term i.e. 11th term is the middle term. Hence, the middle term = $$T_{11}$$ $$T_{11}$$ = $$T_{10 + 1}$$ = $$^{20}C_{10}$$ $$({2\over 3}x^2)^{20 – 10}$$ $$(-{3\over 2x})^{10}$$ = $$^{20}C_{10} x^{10}$$ Similar Questions Find the middle term in the expansion …

Solution : We know that the (r + 1)th term or general term in the expansion of $$(x + a)^n$$ is given by $$T_{r + 1}$$ = $$^{n}C_r x^{n – r} a^r$$ In the expansion of $$({x\over a} – {3a\over x^2})^{12}$$, we have $$T_{9}$$ = $$T_{8 + 1}$$ = $$^{12}C_8 ({x\over a})^{12 – 8} ({-3a\over … ## Find the 10th term in the binomial expansion of \((2x^2 + {1\over x})^{12}$$.

Solution : We know that the (r + 1)th term or general term in the expansion of $$(x + a)^n$$ is given by $$T_{r + 1}$$ = $$^{n}C_r x^{n – r} a^r$$ In the expansion of $$(2x^2 + {1\over x})^{12}$$, we have $$T_{10}$$ = $$T_{9 + 1}$$ = $$^{12}C_9 (2x^2)^{12 – 9} ({1\over x})^9$$ $$\implies$$ …

## Which is larger $$(1.01)^{1000000}$$ or 10,000?

Solution : We have, $$(1.01)^{1000000}$$  – 10000 = $$(1 + 0.01)^{1000000}$$ – 10000 By using binomial theorem, = $$^{1000000}C_0$$ + $$^{1000000}C_1 (0.01)$$  + $$^{1000000}C_2 (0.01)^2$$  + …… + $$^{1000000}C_{1000000} (0.01)^{1000000}$$ – 10000 = (1 + 1000000(0.01) + other positive terms) – 10000 = (1 + 10000 + other positive terms) – 10000 = 1 + …

## By using binomial theorem, expand $$(1 + x + x^2)^3$$.

Solution : Let y = x + $$x^2$$. Then, $$(1 + x + x^2)^3$$ = $$(1 + y)^3$$ = $$^3C_0$$ + $$^3C_1 y$$ + $$^3C_2 y^2$$ + $$^3C_3 y^3$$ = $$1 + 3y + 3y^2 + y^3$$ = 1 + 3$$(x + x^2)$$ + 3$$(x + x^2)^2$$ + $$(x + x^2)^3$$ = \(x^6 + 3x^5 …