THE BINOMIAL THEOREM shows how to calculate a power of a binomial –
(x+ y)n -- without actually multiplying out.
For example, if we actually multiplied out the 4th power of (x + y) --
(x + y)4 = (x + y) (x + y) (x + y) (x + y)
-- then on collecting like terms we would find:
(x + y)4 = x4 + 4x3y + 6x2y2 + 4xy3 + y4 . . . . . (1)
THE BINOMIAL THEOREM shows how to calculate a power of a binomial –
(x+ y)n -- without actually multiplying out.
For example, if we actually multiplied out the 4th power of (x + y) --
(x + y)4 = (x + y) (x + y) (x + y) (x + y)
-- then on collecting like terms we would find:
(x + y)4 = x4 + 4x3y + 6x2y2 + 4xy3 + y4 . . . . . (1)
JEE Mathematics/ Lakshmikanta Satapathy/ Binomial Theorem QA part 5/ JEE Question on finding the sum of all coefficients in a Trinomial expansion solved with the related concepts
Behavior of Trigonometric Ratios in the quadrants , Important Trigonometric Identities involving multiple and compound angles and values of trigonometric ratios for known angles
JEE Mathematics/ Lakshmikanta Satapathy/ Binomial Theorem QA part 5/ JEE Question on finding the sum of all coefficients in a Trinomial expansion solved with the related concepts
Behavior of Trigonometric Ratios in the quadrants , Important Trigonometric Identities involving multiple and compound angles and values of trigonometric ratios for known angles
2D Geometry Distance formula, Slope formula , section formula, Area of triangle formula, angle between straight lines, parallel and perpendicular lines
JEE Mathematics/ Lakshmikanta Satapathy/ Quadratic Equation part 2/ Question on properties of the roots of a quadratic equation solved with the related concepts
JEE Physics/ Lakshmikanta Satapathy/ Question on the magnitude and direction of the resultant of two displacement vectors asked by a student solved in the slides
JEE Mathematics/ Lakshmikanta Satapathy/ Inverse trigonometry QA part 6/ Questions on Inverse trigonometric functions involving tan inverse function solved with the related concepts
JEE Physics/ Lakshmikanta Satapathy/ Laws of Motion QA part 7/ Question on Breaking stress of wire connected in a pulley block system solved with the related concepts
JEE Physics/ Lakshmikanta Satapathy/ Transient current QA part 1/ JEE question on maximum and minimum current from a DC source connected across Inductance and Resistance solved with the related concepts
JEE Mathematics/ Lakshmikanta Satapathy/ Inverse Trigonometry QA part 5/ Question on sin inverse cosine inverse and tan inverse solved with the related concepts
JEE Mathematics/ Lakshmikanta Satapathy/ Binomial theorem part 6/ JEE Question on the coefficient of a given power of x in the expansion of 100 factors
JEE Mathematics/ Lakshmikanta Satapathy/ Probability QA part 12/ JEE Question on Probability involving the complex cube roots of unity is solved with the related concepts
JEE Physics/ Lakshmikanta Satapathy/ Electromagnetism QA part 7/ Question on doubling the range of an ammeter by shunting solved with the related concepts
JEE Mathematics/ Lakshmikanta Satapathy/ Indefinite Integration QA part 21/ Question on Indefinite integration is solved resolving the integrand into partial fractions
JEE Mathematics/ Lakshmikanta Satapathy/ Questions on Indefinite Integration part 12 taken from previous Board papers solve by the method of substitution using standard Integrals
SCRUTINY TO THE NON-AXIALLY DEFORMATIONS OF AN ELASTIC FOUNDATION ON A CYLIND...P singh
This paper is devoted to homogenization of partial differential operators to use in special structure that is a plate allied to an elastic foundation when it is situated through the basic loads (especially with the harmonic forces) with a Non-axially deformation of the cantilever. Furthermore, it contains the Equations of motion that they can be derived from degenerate Non-linear elliptic ones. Through the mentioned processes, there exists many excess works related to computing the bounded conditions for this special application form of study (when the deformation phenomenon has occurred). At the end of the article whole results of the study on a circular plate are debated and new ways assigned to them are discussed. Afterwards all the processes are formulised with the collection of contracting sequences and expanding sequences integrable functions that are intrinsically joints with the characteristic functions to expanding the behaviour of an elastic foundation. Thenceforth all the resultant functions are sets and compared with the other ones (without the loads). Sample pictures and analysis of the study were employed with the ANSYS software to obtain the better observations and conclusions.
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First-order cosmological perturbations produced by point-like masses: all sca...Maxim Eingorn
This presentation based on the paper http://arxiv.org/abs/1509.03835 was made at Institute of Cosmology, Tufts University, on November 12, 2015. The abstract follows:
In the framework of the concordance cosmological model the first-order scalar and vector perturbations of the homogeneous background are derived without any supplementary approximations in addition to the weak gravitational field limit. The sources of these perturbations (inhomogeneities) are presented in the discrete form of a system of separate point-like gravitating masses. The obtained expressions for the metric corrections are valid at all (sub-horizon and super-horizon) scales and converge in all points except the locations of the sources, and their average values are zero (thus, first-order backreaction effects are absent). Both the Minkowski background limit and the Newtonian cosmological approximation are reached under certain well-defined conditions. An important feature of the velocity-independent part of the scalar perturbation is revealed: up to an additive constant it represents a sum of Yukawa potentials produced by inhomogeneities with the same finite time-dependent Yukawa interaction range. The suggesting itself connection between this range and the homogeneity scale is briefly discussed along with other possible physical implications.
JEE Physics/ Lakshmikanta Satapathy/ Work Energy and Power/ Force and Potential energy/ Angular momentum and Speed of Particle/ MCQ one or more correct
JEE Physics/ Lakshmikanta Satapathy/ MCQ On Work Energy Power/ Work-Energy theorem/ Work done by Gravity/ Work done by Air resistance/ Change in Kinetic Energy of body
CBSE Physics/ Lakshmikanta Satapathy/ Electromagnetism QA/ Magnetic field due to circular coil at center & on the axis/ Magnetic field due to Straight conductor/ Magnetic Lorentz force
CBSE Physics/ Lakshmikanta Satapathy/ Amplitudes of Reflected and Transmitted waves/ Sound as Pressure wave/ Speed of sound in Fluids/ Intensity and Loudness of sound
CBSE Physics/ Lakshmikanta Satapathy/ Wave motion/ Vibration of air columns/ Open & closed pipes/ Fundamental frequency & overtones/ End correction/ Resonance tube
CBSE Physics/ Lakshmikanta Satapathy/ Wave Motion Theory/ Reflection of waves/ Traveling and stationary waves/ Nodes and anti-nodes/ Stationary waves in strings/ Laws of transverse vibration of stretched strings
CBSE Physics/ Lakshmikanta Satapathy/ Wave theory/ path difference and Phase difference/ Speed of sound in a gas/ Intensity of wave/ Superposition of waves/ Interference of waves
JEE Mathematics/ Lakshmikanta Satapathy/ Definite integrals part 8/ JEE question on definite integral involving integration by parts solved with complete explanation
JEE Physics/ Lakshmikanta Satapathy/ Alternating Current Theory part 5/ Theory of Resonance in LCR Series circuit with Sharpness of resonance Half power Band width and Quality factor
JEE Mathematics/ Lakshmikanta Satapathy/ Set theory part 3/ JEE question on Power set of the cartesian product of two sets solved with the related concepts
JEE Mathematics/ Lakshmikanta Satapathy/ Permutation and Combination QA part 3/ JEE Question on selection of team from club members with boy and girl combination solved with the related concepts
JEE Physics/ Lakshmikanta Satapathy/ Vectors QA part 1/ Question on rectangular components of vectors asked by a student solved with the related concepts
JEE Mathematics/ Lakshmikanta Satapathy/ Permutation and Combination QA part 2/ JEE question on four digit numbers permuted from five given digits solved with the related concepts
JEE Physics/ Lakshmikanta Satapathy/ Alternating Current Theory Part 4/ Phase relationship between current and voltage in a series LCR circuit discussed in details
The French Revolution, which began in 1789, was a period of radical social and political upheaval in France. It marked the decline of absolute monarchies, the rise of secular and democratic republics, and the eventual rise of Napoleon Bonaparte. This revolutionary period is crucial in understanding the transition from feudalism to modernity in Europe.
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A Strategic Approach: GenAI in EducationPeter Windle
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2. Binomial Theorem 1
Physics Helpline
L K Satapathy
1 2 2
0 1 2( ) ...n n n n n n n
a b C a C a b C a b
1 2 2
0 1 2( ) ...n n n n n n n
a b C a C a b C a b
Subscript of C increases from 0 to n
Power of a decreases from n to 0
Power of b increases from 0 to n
Sum of the powers of a and b = n
...n n r r n n
r nC a b C b
( 1) .. ( 1)r n n r r n n n
r nC a b C b
3. Binomial Theorem 1
Physics Helpline
L K Satapathy
2
0 1 2(1 ) ...n n n n n n
nx C C x C x C x
2
0 1 2(1 ) .. ( 1)n n n n n n n
nx C C x C x C x
0 1 21 (1 1) ...n n n n n
nx C C C C
0 1 2 3 41 0 . . .n n n n n
x C C C C C
0 2 4 1 3 5( . . . ) ( . . . )n n n n n n
C C C C C C
Sum of odd coefficients = Sum of even coefficients
0
2
n
n n
r
r
C
4. Binomial Theorem 1
Physics Helpline
L K Satapathy
1 2 2
0 1 2( ) ...n n n n n n n n n
na b C a C a b C a b C b
( r + 1)th term 1 . .n n r r
r rt C a b
Subscript of C = r
Power of a = n – r
Power of b = r
1st term = 0
0. .n n
C a b
2nd term = 1 1
1. .n n
C a b
3rd term = 2 2
2. .n n
C a b
The General Term
5. Binomial Theorem 1
Physics Helpline
L K Satapathy
m terms on either side of the middle term
( 1) 1
2
th
th nThe middle term m term term
Middle Term (s)
1 2 1 ( )Number of terms n m odd
Power of (a + x) = n Number of terms = n +1
When n is even : 2
2
nWe put n m m
1We get middle term
Illustration
n = 12 13 terms : 1 2 3 4 5 6 7 8 9 10 11 12 13
6 Terms
on Left
6 Terms
on RightMiddle term
6. Binomial Theorem 1
Physics Helpline
L K Satapathy
2 ( 1) ( 2)th th
The middle terms m and m terms
When n is odd : 12 1
2
nWe put n m m
1 2 2 ( )Number of terms n m even
2We get middle terms
2m terms on either side of the middle terms
1 3,
2 2
th th
n n terms
Illustration
n = 13 14 terms : 1 2 3 4 5 6 7 8 9 10 11 12 13 14
6 Terms
on Left
6 Terms
on Right2 Middle terms
7. Physics Helpline
L K Satapathy
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