Sigma I Eq 1 To N Sigma J Eq 1 To I Sigma K Eq 1 To J 1 Is Equal To
Solution: ∑i = 1n ∑j = 1i∑k= 1j 1 = ∑i = 1n ∑j = 1i j = ∑i =... View Article
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Sum Of N Terms Of The Following Series 1 3 3 3 5 3 7 3 Is
Solution: 13+33+53+73+….+n3 = ∑r=1n(2r-1)3 = ∑r=1n(8r3-12r2+6r-1) = 8∑r=1nr3 -12∑r=1n r2+6∑r=1n r-∑r=1n1 = 8[(n(n+1))2/4]-12[n(n+1)(2n+1)/6]+[6(n(n+1)/2]-n = 2n2(n+1)2-2n(n+1)(2n+1)+3n(n+1)-n = 2n(n+1)[n(n+1)-2n-1]+n[3n+3-1] = 2n(n+1)(n2-n-1)+n(3n+2) =... View Article
If Hn 1 1 2 1 N Then The Value Of Sn 1 3 2 5 3 Is
If Hn = 1 + 1/2 + …+ 1/n, then the value of Sn = 1 + 3/2 + 5/3... View Article
If N Is Odd Positive Integer And 1 X X 2 X 3 N Sigma R 0 To 3n Ar Xr Then A0 A1 A2 A3 A3n Is Equal To
If n is odd positive integer and (1+x+x^2+x^3)^n = sigma r = 0 to 3n arx^r, then a0-a1+a2-a3+…-a3n is equal... View Article
The value of 1+(1/3)+(1.3)/(3.6) + (1.3.5)/(3.6.9)+(1.3.5.7)/(3.6.9.12) +…. is equal to
(1) \(\begin{array}{l}\sqrt{2}\end{array} \) (2) \(\begin{array}{l}\sqrt{3}\end{array} \) (3) \(\begin{array}{l}\sqrt{\frac{3}{2}}\end{array} \) (4) \(\begin{array}{l}\sqrt{\frac{1}{2}}\end{array} \) Solution: 1+(⅓)+(1.3)/(3.6) + (1.3.5)/(3.6.9)+ (1.3.5.7)/(3.6.9.12) +… Let y... View Article
Let The Positive Numbers A B C D Be In Ap Then Abc Abd Acd Bcd Are
Solution: Given a,b,c,d are in AP. So d,c,b,a are in AP. 1/d, 1/c, 1/b, 1/a are in HP. So abcd/d,... View Article
If Sn Sigma K 1 To 4n 1 K K 1 2k 2 Then Sn Can Take Values
Solution: ∑k=14n (-1)k(k+1)/2k2 = -12-22+32+42-52-62+72+82-….+(4n)2 = (-12+32-52+72-…+(4n-1)2)+(-22+42-62+82-..+(4n2) = 2(4+12+20+…(8n-4))+2(6+14+22+…(8n-2)) ( Sum of n terms in AP) = n(8n)+n(8n+4)… (we use... View Article
Let Alpha Beta Be The Roots Of X2 X P 0 And Gamma And Delta Be Roots Of X2 4x Q 0 If Alpha Beta Gamma Delta Are In Gp Then Integral Values Of P And Q Respectively Are
Solution: Given α, β are the roots of x2-x+p = 0 Sum of roots, α+β = 1…(i) Product of roots,... View Article
Sigma K 1 To 2n 1 1 K 1 K2 Equals
Solution: ∑k=12n+1(-1)k-1k2= 12-22+32-42+…(2n)2+(2n+1)2 = [12+32+52+…(2n+1)2]-[22+42+62+…+(2n)2] = [12+22+32+…(2n+1)2]-2[22+42+62+…(2n)2] = ∑(2n+1)2-2(22)[1+22+32+…n2] …(1) We know ∑n2 = n(n+1)(2n+1)/6 Equation (1) becomes (2n+1)(2n+1+1)(2(2n+1)+1)-8[n(n+1)(2n+1)/6] =... View Article
Let Alpha Beta Denote The Cube Roots Of Unity Other Than 1 And Alpha Not Equal Beta If S Sigma Then The Value Of S Is
Solution: S = ∑n=0302 (-1)n(α/ β)n Cube roots of unity are 1, ω, ω2. Take β = ω and α... View Article
The Sum Of The Series Sigma N 8 To 17 1 N 2 N 3 Is Equal To
Solution: S = ∑n=817 1/(n+2)(n+3) = ∑n=817 1/(n+2) -1/(n+3) = ((1/10)-(1/11)+(1/11)-(1/12)+…+(1/19)-(1/20)) = (1/10)-(1/20) = 1/20 Hence option (4) is the... View Article
The Value Of 2 Power 1 By 4 4 Power 1 By 8 Infinity Is
Solution: 21/4 × 41/8 ×81/16 … ∞ S = (1/4)+(2/8)+(3/16)+… Divide by 2 S/2 = (1/8)+(2/16)+(3/32)+… Subtracting S/2 = (1/4)+(1/8)+(1/16)+…... View Article
The Sum Of The Series 1 3x 6×2 10×3 Infinity Will Be
Solution: Let S = 1+3x+6x2+10x3 + …∞ ..(i) Multiply by x xS = x+3x2+6x3+10x4+ …∞ (ii) (i)-(ii) (1-x)S = 1+2x+3x2+4x3+….∞... View Article
If Y 3x 1 3 X 1 X Real Then The Least Value Of Y Is
Solution: Given y = 3x-1 + 3– x-1 = (3x/3)+(1/3x×3) We know AM ≥GM ((3x/3)+(3/3x))/2 ≥ √(3x/3)(1/3×3x) ((3x/3)+(3/3x))/2 ≥ √(1/9)... View Article
Let A B C Be In Ap And A 1 B 1 C 1 If X 1 A A2 Infinity Y 1 Then X Y And Z Are In
Solution: Since a,b,c in AP 2b = a+c x = 1 + a + a2 + …+ ∞ This is... View Article
If Fx Cos 2x Plus Sec 2x Then
Solution: We know AM ≥ GM (cos2x+sec2x )/2 ≥ √(cos2x sec2x) (cos2x+sec2x ) ≥ 2√(cos2x (1/cos2x) f(x) ≥ 2 Hence... View Article
If A1 A2 G1 G2 And H1 H2 Are Two Am Gm And Hm Between Two Quantities Then The Value Of
If A1, A2, G1 , G2 and H1, H2 are two AM’s, GM’s and HM’s between two quantities, then the... View Article
If H1 And H2 Are Two Harmonic Means Between Two Positive Numbers A And B A Not Equal B A And G Are The Arithmetic And Geometric Means Between A And B Then H2 Plus H1 By H1h2 Is
If H1 and H2 are two harmonic means between two positive numbers a and b (a not equal b), A... View Article
If Am Of Two Numbers Is Twice Of Their Gm Then The Ratio Of Greatest Number To Smallest Number Is
Solution: Let a and b be the two numbers. AM = 2 GM (a+b)/2 = 2√(ab) (a+b) = 4√(ab) Squaring... View Article
Three Numbers Whose Sum Is 15 Are In Ap If They Are Added By 1 4 And 19 Respectively Then They Are In Gp The Numbers Are
Solution: Let a-d,a,a+d are the three numbers in AP. Given sum = 15 a-d+a+a+d = 15 3a = 15 a... View Article
