Question

If ${{\left( 1+x \right)}^{n}}={{P}_{0}}+{{P}_{\grave{\ }}}_{1}x+{{P}_{2}}{{x}^{2}}+{{P}_{3}}{{x}^{3}}+{{P}_{4}}{{x}^{4}}+.........{{P}_{n}}{{x}^{n}}$, ${{P}_{1}}-{{P}_{3}}+{{P}_{5}}-........={{2}^{\dfrac{n}{2}}}\sin \left( \dfrac{n\pi }{m} \right)$ then find m:

Answer 1

Hint: Put value of x as $i=\sqrt{-1}$ to the given expression in the problem. Properties of ‘I’ are given as $i=\sqrt{-1},{{i}^{2}}=-1,{{i}^{3}}=-i,{{i}^{4}}=1$ conversion of any complex number to parametric form is done by dividing and multiplying the complex number by the modules of it. Demovier’s theorem is given as ${{\left( \cos \theta +i\sin \theta \right)}^{n}}=\left( \operatorname{cosn}\theta +i\operatorname{sinn}\theta \right)$.

Complete step-by-step answer:
We are given expansion of ${{\left( 1+x \right)}^{n}}$as
${{\left( 1+x \right)}^{n}}={{P}_{0}}+{{P}_{\grave{\ }}}_{1}x+{{P}_{2}}{{x}^{2}}+{{P}_{3}}{{x}^{3}}+{{P}_{4}}{{x}^{4}}+.........{{P}_{n}}{{x}^{n}}$ …………. (i)
And we need to determine the value of ‘m’ if the equation given below is true.
${{P}_{1}}-{{P}_{3}}+{{P}_{5}}-{{P}_{7}}...........={{2}^{\dfrac{n}{2}}}\sin \left( \dfrac{n\pi }{m} \right)$ ………….. (ii)
So, we have expansion of ${{\left( 1+x \right)}^{n}}$as
${{\left( 1+x \right)}^{n}}={{P}_{0}}+{{P}_{\grave{\ }}}_{1}x+{{P}_{2}}{{x}^{2}}+{{P}_{3}}{{x}^{3}}+{{P}_{4}}{{x}^{4}}+.........+{{P}_{n}}{{x}^{n}}$
Put the value of $x=i=\sqrt{-1}$. Hence, we get
${{\left( 1+i \right)}^{n}}={{P}_{0}}+{{P}_{\grave{\ }}}_{1}\left( i \right)+{{P}_{2}}{{\left( i \right)}^{2}}+{{P}_{3}}{{\left( i \right)}^{3}}+{{P}_{4}}{{\left( i \right)}^{4}}+{{P}_{5}}{{\left( i \right)}^{5}}+{{P}_{6}}{{\left( i \right)}^{6}}+.........+{{P}_{n}}{{\left( i \right)}^{n}}$
Now, we know the property of iota i.e. as
$i=\sqrt{-1},{{i}^{2}}=-1,{{i}^{3}}=-i,{{i}^{4}}=1$
Hence, we can write the expansion of ${{\left( 1+x \right)}^{n}}$with the help of above properties of ‘i’ as
${{\left( 1+i \right)}^{n}}={{P}_{0}}+i{{P}_{1}}-{{P}_{2}}-i{{P}_{3}}+{{P}_{4}}+i{{P}_{5}}-{{P}_{6}}+.........+......$
Hence, we can write the R.H.S. of the above expression of the form of a + ib as
${{\left( 1+i \right)}^{n}}=\left( {{P}_{0}}-{{P}_{2}}+{{P}_{4}}-{{P}_{6}}+....... \right)+i\left( {{P}_{1}}-{{P}_{3}}+{{P}_{5}}-{{P}_{7}}+....... \right)$ …………… (iii)
Now, we can convert (1 + i) to polar form of complex number by dividing and multiplying (1 + i) by the modulus of this complex number. As, we know modules of any complex number (x + iy) is given as
$\left| x+iy \right|=\sqrt{{{x}^{2}}+{{y}^{2}}}$
Hence, modules of (I + i) is given as
$=\sqrt{{{\left( 1 \right)}^{2}}+{{\left( 1 \right)}^{2}}}=\sqrt{2}$
Now, let us represent (1 + i) by z. so, we get
z = (1 + i)
Now, multiply (1 + i) by $\sqrt{2}$ and divide it as well. So, we get
$z=\sqrt{2\left( \dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}i \right)}$
We know $\cos \dfrac{\pi }{4}=\sin \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}}$.
Hence, we can replace $\dfrac{1}{\sqrt{2}}$in the expression $z=\sqrt{2\left( \dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}i \right)}$ by $\cos \dfrac{\pi }{4},\sin \dfrac{\pi }{4}$. So, we get
$z=\sqrt{2}\left( \cos \dfrac{\pi }{4}+i\sin \dfrac{\pi }{4} \right)$……………. (iv)
Now, we can replace (1 + i) in the equation (iii) by the polar form of it, which is given in the equation (iv). So, we get
${{\left( \sqrt{2}\left( \cos \dfrac{\pi }{4}+i\sin \dfrac{\pi }{4} \right) \right)}^{n}}=\left( {{P}_{0}}-{{P}_{2}}+{{P}_{4}}-{{P}_{6}}+....... \right)+i\left( {{P}_{1}}-{{P}_{3}}+{{P}_{5}}-{{P}_{7}}+....... \right)$
We know ${{\left( ab \right)}^{n}}={{a}^{n}}{{b}^{n}}$. So, we get
${{\left( \sqrt{2} \right)}^{n}}{{\left( \cos \dfrac{\pi }{4}+i\sin \dfrac{\pi }{4} \right)}^{n}}=\left( {{P}_{0}}-{{P}_{2}}+{{P}_{4}}-{{P}_{6}}+....... \right)+i\left( {{P}_{1}}-{{P}_{3}}+{{P}_{5}}-{{P}_{7}}+....... \right)$ ……….. (v)
Now, we know the demoveir’s theorem can be given as
${{\left( \cos \theta +i\sin \theta \right)}^{n}}=\cos n\theta +i\operatorname{sinn}\theta$ ……………… (vi)
Hence, we can use the above theorem to simplify the expression in the left hand side of the equation (v). so, we get
${{\left( \sqrt{2} \right)}^{n}}\left( \cos \left( \dfrac{n\pi }{4} \right)+i\sin \left( \dfrac{n\pi }{4} \right) \right)=\left( {{P}_{0}}-{{P}_{2}}+{{P}_{4}}-{{P}_{6}}+....... \right)+i\left( {{P}_{1}}-{{P}_{3}}+{{P}_{5}}-{{P}_{7}}+....... \right)$
We can write ${{\left( \sqrt{2} \right)}^{n}}\to {{\left( 2 \right)}^{\dfrac{n}{2}}}$as well.
So, we get
$\left( {{2}^{\dfrac{n}{2}}}\cos \left( \dfrac{n\pi }{4} \right) \right)+i\left( {{2}^{\dfrac{n}{2}}}\sin \left( \dfrac{n\pi }{4} \right) \right)=\left( {{P}_{0}}-{{P}_{2}}+{{P}_{4}}-{{P}_{6}}+....... \right)+i\left( {{P}_{1}}-{{P}_{3}}+{{P}_{5}}-{{P}_{7}}+....... \right)$ …(vii)
Now, we know that two complexes can only be equal if the real parts and imaginary parts are equal i.e.
If ${{\alpha }_{1}}+i{{\beta }_{1}}={{\alpha }_{2}}+i{{\beta }_{2}}$ then
${{\alpha }_{1}}={{\alpha }_{2}},{{\beta }_{1}}={{\beta }_{2}}$
Hence, on comparing the real and imaginary parts of both sides of equation (vii). So, we get
$\left( {{P}_{0}}-{{P}_{2}}+{{P}_{4}}-{{P}_{6}}+....... \right)={{2}^{\dfrac{n}{2}}}\cos \left( \dfrac{n\pi }{4} \right)$ …………….. (viii)
$\left( {{P}_{1}}-{{P}_{3}}+{{P}_{5}}-{{P}_{7}}+....... \right)={{2}^{\dfrac{n}{2}}}sin\left( \dfrac{n\pi }{4} \right)$ ……………..(ix)
Now, we can compare the equation (ix) and the equation (ii), we get that the R.H.S. of both them will be equal. Hence, we get
$sin\left( \dfrac{n\pi }{4} \right)=sin\left( \dfrac{n\pi }{m} \right)$
So, $\dfrac{n\pi }{4}=\dfrac{n\pi }{m}$
m = 4
Therefore, the value of m is 4.

Note:One may go wrong if he/she tries to find values of ${{P}_{0}},{{P}_{1}},{{P}_{2}},{{P}_{3}}{{P}_{4}}..........{{P}_{n}}$ , then try to get the required answer; so, one will get values of ${{P}_{0}},{{P}_{1}},{{P}_{2}},{{P}_{3}}.....$ as 1, $C_{1}^{n},C_{2}^{n},C_{3}^{n}........$ etc.
But it will not help to get the required answer. So, just observe the given relation only.
We don’t need to write the end term of the series found by putting x = I, as last term will depend on the value of ‘n’. so, don’t confuse this part of the solution.
Put x = I, using Demoivre's theorem are the key points of the question.