ABCD is a square of length a, `a in N`, a > 1. Let `L_1, L_2 , L_3...` be points on BC such that `BL_1 = L_1 L_2 = L_2 L_3 = .... 1` and `M_1,M_2 , M_3,....`be points on CD such that `CM_1 = M_1M_2= M_2 M_3=... = 1`. Then `sum_(n = 1)^(a-1) ((AL_n)^2 + (L_n M_n)^2)` is equal to :

sum_(m=1)^(n) tan^(-1) ((2m)/(m^(4) + m^(2) + 2)) is equal to

ABC is a right angled triangle in which angleB=90^@ and BC=a. If n points L_1,L_2,....L_n on AB are such that AB is divided in n+1 equal parts and L_1M_1,L_2M_2,...L_nM_n are line segments parallel to BC and M_1,M_2,...M-n are on AC. Then the sum og the lengths of L_1M_1,L_2M_2,.....L_nM_n is

A B C is a right-angled triangle in which /_B=90^0 and B C=adot If n points L_1, L_2, ,L_nonA B is divided in n+1 equal parts and L_1M_1, L_2M_2, ,L_n M_n are line segments parallel to B Ca n dM_1, M_2, ,M_n are on A C , then the sum of the lengths of L_1M_1, L_2M_2, ,L_n M_n is (a(n+1))/2 b. (a(n-1))/2 c. (a n)/2 d. none of these

ABC is a right angled Delta in which angleB=90^@ and BC =a. If n points L_1,L_2,L_3,...,L_n on AB are such that AB is divided into (n+1) equal parts and L_1M_1,L_2M_2,...,L_nM_n are line segments parallel to BC and points M_1,M_2,...,M_n are on AC, then the sum of the lengths of L_1M_1,L_2M_2,...,L_nM_n is:

Let a_1,a_2 ,a_3 ...... be an A.P. Prove that sum_(n=1)^(2m)(-1)^(n-1)a_n^2=m/(2m-1)(a_1^2-a_(2m)^2) .

What is shape of the orbital with (i) n = 2 and l = 0 , (ii) n = 2 and l = 1?

Prove that ^mC_1^n C_m-^m C_2^(2n)C_m+^m C_3^(3n)C_m-.....=(-1)^(m-1)n^mdot

The value of .^(n)C_(1)+.^(n+1)C_(2)+.^(n+2)C_(3)+"….."+.^(n+m-1)C_(m) is equal to

If m = sqrtfrac(n)(n+1/2) and m = 1/2, then n =