QuestionsQuestion 1Expand the following upto 4 terms, taking the values of xxx such that the expansion in each case is valid:(i)(1+x)−1/3(1+x)^{-1/3}(1+x)−1/3(ii)(4−3x)1/2(4-3x)^{1/2}(4−3x)1/2(iii)(1−x)−1(1+x)2\dfrac{(1-x)^{-1}}{(1+x)^2}(1+x)2(1−x)−1(iv)1+2x1−x\dfrac{\sqrt{1+2x}}{1-x}1−x1+2xSolutionTheoryQuestion 2Find the coefficient of xnx^nxn in the expansion of:(i)1+x2(1+x)2\dfrac{1+x^2}{(1+x)^2}(1+x)21+x2(ii)(1+x)2(1−x)2\dfrac{(1+x)^2}{(1-x)^2}(1−x)2(1+x)2SolutionTheoryQuestion 3If xxx is so small that its square and higher powers can be neglected, then show that:(i)1−x1+x≈1−32x\dfrac{1-x}{\sqrt{1+x}} \approx 1 - \dfrac{3}{2}x1+x1−x≈1−23x(ii)1+2x1−x≈1+32x\dfrac{\sqrt{1+2x}}{\sqrt{1-x}} \approx 1 + \dfrac{3}{2}x1−x1+2x≈1+23x(iii)(9+7x)1/2−(16+3x)1/44+5x≈14−17384x\dfrac{(9+7x)^{1/2} - (16+3x)^{1/4}}{4+5x} \approx \dfrac{1}{4} - \dfrac{17}{384}x4+5x(9+7x)1/2−(16+3x)1/4≈41−38417x(iv)4+x(1−x)3≈2+254x\dfrac{\sqrt{4+x}}{(1-x)^3} \approx 2 + \dfrac{25}{4}x(1−x)34+x≈2+425xSolutionTheoryQuestion 4If xxx is so small that its cube and higher power can be neglected, show that:(i)1−x−2x2≈1−12x−98x2\sqrt{1-x-2x^2} \approx 1 - \dfrac{1}{2}x - \dfrac{9}{8}x^21−x−2x2≈1−21x−89x2(ii)1+x1−x≈1+x+12x2\sqrt{\dfrac{1+x}{1-x}} \approx 1 + x + \dfrac{1}{2}x^21−x1+x≈1+x+21x2SolutionTheoryQuestion 5If xxx is very nearly equal to 1, then prove that pxp−qxq≈(p−q)xp+qpx^p - qx^q \approx (p-q)x^{p+q}pxp−qxq≈(p−q)xp+q.SolutionTheoryQuestion 6Identify the following series as binomial expansion and find the sum. 1−12(14)+1⋅32!(14)2−1⋅3⋅53!(14)3+⋯1 - \dfrac{1}{2}\left(\dfrac{1}{4}\right) + \dfrac{1\cdot 3}{2!}\left(\dfrac{1}{4}\right)^2 - \dfrac{1\cdot 3\cdot 5}{3!}\left(\dfrac{1}{4}\right)^3 + \cdots1−21(41)+2!1⋅3(41)2−3!1⋅3⋅5(41)3+⋯SolutionTheoryQuestion 7Use binomial theorem to show that 1+14+1⋅34⋅8+1⋅3⋅54⋅8⋅12+⋯=21 + \dfrac{1}{4} + \dfrac{1\cdot 3}{4\cdot 8} + \dfrac{1\cdot 3\cdot 5}{4\cdot 8\cdot 12} + \cdots = \sqrt{2}1+41+4⋅81⋅3+4⋅8⋅121⋅3⋅5+⋯=2.SolutionTheoryQuestion 8If y=13+1⋅32!(13)2+1⋅3⋅53!(13)3+⋯y = \dfrac{1}{3} + \dfrac{1\cdot 3}{2!}\left(\dfrac{1}{3}\right)^2 + \dfrac{1\cdot 3\cdot 5}{3!}\left(\dfrac{1}{3}\right)^3 + \cdotsy=31+2!1⋅3(31)2+3!1⋅3⋅5(31)3+⋯ prove that y2+2y−2=0y^2 + 2y - 2 = 0y2+2y−2=0.SolutionTheoryQuestion 9If 2y=122+1⋅32!⋅124+1⋅3⋅53!⋅126+⋯2y = \dfrac{1}{2^2} + \dfrac{1\cdot 3}{2!}\cdot \dfrac{1}{2^4} + \dfrac{1\cdot 3\cdot 5}{3!}\cdot \dfrac{1}{2^6} + \cdots2y=221+2!1⋅3⋅241+3!1⋅3⋅5⋅261+⋯, prove that 4y2+4y−1=04y^2 + 4y - 1 = 04y2+4y−1=0.SolutionTheoryQuestion 10Show that the coefficient of xrx^rxr in x(1−px)(1−qx)\dfrac{x}{(1-px)(1-qx)}(1−px)(1−qx)x is pr−qrp−q\dfrac{p^r - q^r}{p-q}p−qpr−qr.SolutionTheory