if (1+x)10=a0+a1x+....a10x10, then (a0−a2+a4−a6+a8−a10)2+(a1−a – InterviewSolution

InterviewSolution

1.	if (1+x)10=a0+a1x+....a10x10, then (a0−a2+a4−a6+a8−a10)2+(a1−a3+a5−a7+a9)2 is equal to
Answer» $i f (1 + x)^{10} = a_{0} + a_{1} x + . . . . a_{10} x^{10}, t h e n (a_{0} - a_{2} + a_{4} - a_{6} + a_{8} - a_{10})^{2} + (a_{1} - a_{3} + a_{5} - a_{7} + a_{9})^{2}$ is equal to

Discussion

No Comment Found

Related InterviewSolutions

sin−1[x√1−x−√x√1−x2]=
Let f and g be real functions defined by f(x)=√x+2 and g(x)=√4−x2. Then,the domain of the function (fg)(x) is
The coordinates of the point which divides the line segment joining the points (1, –2, 3) and (3, 4, –5) internally in the ratio 2:3 is (x, y, z). Find the value of x+y+z___
If a1,a2,a3,....,an be an AP of non-zero terms. Prove that 1a1a2+1a2a3+....+1an−1an= n−1a1an
Find the value of C20+C21+C22.....C2n, where Cr=nCr
Match the following complex number with their arguements A––Principal Argument–––––––––––––––––––––––1)5+10ia)120∘2)√3+3ib)−120∘ 3)√12−2ic)tan−12 4)−2−2id)−45∘e)−30∘f)−135∘
A square with each side equal to a lies above the x-axis and has one vertex at the origin. One of the sides passing through the origin makes an angle α(0<α<π4) with the positive direction of the x-axis. Equation of a diagonal of the square is
Four identical particles each of mass 1 kg are arranged at the corners of a square of side length 2Ö2 m. If one of the particles is removed, the shift in the centre of mass will be
The range of f(x)=ex1+[x], x≥0 is
Find the ratio in which the line segment joining the points (4,8,10) and (6,10,-8) is divided by the YZ-plane.