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By @bollywood_new_moviess


https://t.me/latest_hindi_songs

Solve plzzz 🙏🙏🙏 @imgroooot

Let s = 1 + 2x + 3x² + 4x³ + ....

Multiply by x and subtract

s(1 - x) = 1 + x + x² + x³ + ..... = 1/(1 - x)

s = 1/(1 - x)²

Given series = (s)^1/2 = 1/(1 - x)

(s)^1/2 = 1 + x + x³ + x⁴ + .....

Coefficient of x⁴ = 1

Solution please

AM ≥ GM

(x + y + z)/3 ≥ (xyz)^1/3

1/27 ≥ xyz


Again

[(1 + x) + (1 + y) + (1 + z)]/3 ≥ [(1 + x)(1 + y)(1 + z)]^1/3


64/27 ≥ (1 + x)(1 + y)(1 + z)

Using both

64 ≥ (1 + x)(1 + y)(1 + z)/xyz

64

Photo from shivam Jaiswal

Let p = 3, q = 5

j = 1(for first sum)

j = 2(for second sum)

Put values

[5/3] + [3/5] + [6/5]

1 + 0 + 1 = 2


Check RHS = 2*4/4 = 2

LHS = RHS

Plz solve

Put value of tan3x

(3 - tan²x)/(1 - 3tan²x) = p

tan²x = (p - 3)/(3p - 1)

tan²x > 0

(p - 3)/(3p - 1) > 0

p belongs to R - [1/3, 3]

Any short method?

Common multiples of 424, 280 = 2,4,8

Check for 8

424/8 = 53

280/8 = 35 satisfied

Sum of digits = 8

circle : x² + y² = r²

Point on circle

r = 1

P(√2, 0), Q(0, √2)

Required area = Area of triangle - Area of circle part

= 1/2 * √2 * √2 - π/4

= 1 - π/4

Solve

(631)³ * (760)^99 * 256 % 23

(10)³ * (1)^99 * 3 % 23

11 * 1 * 3 % 23

33 % 23

10