Nth power of a real number

Nth power of a number : Some more examples
We know that
(x+a)^n=∑_(k=0)^n▒〖(n¦k) x^k a^(n-k) 〗
Now, we use the above theorem to find out the 8th power of 16 in the following manner,
〖16〗^(8 )=(1^(8 ) ) (8×1^(7 )×6^(1 ) ) (28×1^6×6^2 ) (56×1^5×6^3 ) (70×1^4×6^4 ) (56×1^3×6^5 ) (28×1^2×6^6 ) (8×1^1×6^7 ) (6^8)
= (1) (48) (1008) (12096) (90720) (435456) (1306368) (2239488) (1679616)
= (1) (48) (1008) (12096) (90720) (435456) (1306368) (2239488+167961) (6)
= (1) (48) (1008) (12096) (90720) (435456) (1306368) (2407449) (6)
= (1) (48) (1008) (12096) (90720) (435456) (1306368+240744) (9) (6)
= (1) (48) (1008) (12096) (90720) (435456) (1547112) (9) (6)
= (1) (48) (1008) (12096) (90720) (435456+154711) (2) (9) (6)
= (1) (48) (1008) (12096) (90720) (590167) (2) (9) (6)
= (1) (48) (1008) (12096) (90720+59016) (7) (2) (9) (6)
= (1) (48) (1008) (12096) (149736) (7) (2) (9) (6)
= (1) (48) (1008) (12096+14973) (6) (7) (2) (9) (6)
= (1) (48) (1008) (27069) (6) (7) (2) (9) (6)
= (1) (48) (1008+2706) (9) (6) (7) (2) (9) (6)
= (1) (48) (3714) (9) (6) (7) (2) (9) (6)
= (1) (48+371) (4) (9) (6) (7) (2) (9) (6)
= (1) (419) (4) (9) (6) (7) (2) (9) (6)
= (1+41) (9) (4) (9) (6) (7) (2) (9) (6)
= (42) (9) (4) (9) (6) (7) (2) (9) (6)
= 4294967296. Hence the result.

Again, take another example.
〖38〗^(8 )=(3^(8 ) ) ( 8×3^(7 )×8^1 ) (28×3^(6 )×8^(2 ) ) (56×3^(5 )×8^(3 ) ) (70×3^(4 )×8^(4 ) )(56×3^(3 )×8^(5 ) ) (28×3^(2 )×8^(6 ) ) (8×3^(1 )×8^(7 ) ) (8^(8))
= (6561) (139968) (1306368) (6967296) (23224320) (49545216) (66060288) (50331648) (16777216)
= (6561) (139968) (1306368) (6967296) (23224320) (49545216) (66060288) (50331648+1677721) (6)
= (6561) (139968) (1306368) (6967296) (23224320) (49545216) (66060288) (52009369) (6)
= (6561) (139968) (1306368) (6967296) (23224320) (49545216) (66060288+5200936) (9) (6)
= (6561) (139968) (1306368) (6967296) (23224320) (49545216) (71261224) (9) (6)
= (6561) (139968) (1306368) (6967296) (23224320) (49545216+7126122) (4) (9) (6)
= (6561) (139968) (1306368) (6967296) (23224320) (56671338) (4) (9) (6)
= (6561) (139968) (1306368) (6967296) (23224320+5667133) (8) (4) (9) (6)
= (6561) (139968) (1306368) (6967296) (28891453) (8) (4) (9) (6)
= (6561) (139968) (1306368) (6967296+2889145) (3) (8) (4) (9) (6)
= (6561) (139968) (1306368) (9856441) (3) (8) (4) (9) (6)
= (6561) (139968) (1306368+985644) (1) (3) (8) (4) (9) (6)
= (6561) (139968) (2292012) (1) (3) (8) (4) (9) (6)
= (6561) (139968+229201) (2) (1) (3) (8) (4) (9) (6)
= (6561) (369169) (2) (1) (3) (8) (4) (9) (6)
= (6561+36916) (9) (2) (1) (3) (8) (4) (9) (6)
= (43477) (9) (2) (1) (3) (8) (4) (9) (6)
= 4347792138496. Hence the result.

Now look at the 8th power of 83 and 61.

〖83〗^8=(8^8 ) (8×8^7×3^1 ) (28×8^6×3^2 ) (56×8^(5 )×3^3 ) (70×8^4×3^4 ) (56×8^3×3^5 ) (28×8^2×3^6 ) (8×8^1×3^7 ) (3^8)
= (16777216) (50331648) (66060288) (49545216) (23224320) (6967296) (1306368) (139968) (6561)
Repeat the steps as above; we get 8th power of 83
= 2252292232139041.

Try it for 〖61〗^8 and for some other numbers.

Square

Square of a number x1x2x3...xr−1xr.
Take a number 26374.
Express it as x1x2x3x4x5.
Then square of x1x2x3x4x5 is equal to

In this manner you can find out square of any number.
Some more examples.
Find square of 45128 as shown below

The above series is the expansion of the basic postulates, 

 (xy)^2 = x(x+1)y^2 + 20x(y-5).

Look at the following examples as shown below.

Nth power of a number : some more example

Nth power of a number : Some more examples

We know that

(x+a)^n=∑_(k=0)^n▒〖(n¦k) x^k a^(n-k) 〗

Now, we use the above theorem to find out the 8th power of 16 in the following manner,

〖16〗^(8 )=(1^8) (8×1^7×6^1 ) (28×1^6×6^2 ) (56×1^5×6^3 ) (70×1^4×6^4 ) (56×1^3×6^5 ) (28×1^2×6^6 ) (8×1^1×6^7 ) (6^8)

= (1) (48) (1008) (12096) (90720) (435456) (1306368) (2239488) (1679616)
= (1) (48) (1008) (12096) (90720) (435456) (1306368) (2239488+167961) (6)
= (1) (48) (1008) (12096) (90720) (435456) (1306368) (2407449) (6)
= (1) (48) (1008) (12096) (90720) (435456) (1306368+240744) (9) (6)
= (1) (48) (1008) (12096) (90720) (435456) (1547112) (9) (6)
= (1) (48) (1008) (12096) (90720) (435456+154711) (2) (9) (6)
= (1) (48) (1008) (12096) (90720) (590167) (2) (9) (6)
= (1) (48) (1008) (12096) (90720+59016) (7) (2) (9) (6)
= (1) (48) (1008) (12096) (149736) (7) (2) (9) (6)
= (1) (48) (1008) (12096+14973) (6) (7) (2) (9) (6)
= (1) (48) (1008) (27069) (6) (7) (2) (9) (6)
= (1) (48) (1008+2706) (9) (6) (7) (2) (9) (6)
= (1) (48) (3714) (9) (6) (7) (2) (9) (6)
= (1) (48+371) (4) (9) (6) (7) (2) (9) (6)
= (1) (419) (4) (9) (6) (7) (2) (9) (6)
= (1+41) (9) (4) (9) (6) (7) (2) (9) (6)
= (42) (9) (4) (9) (6) (7) (2) (9) (6)
= 4294967296. Hence the result.

Again, take another example.

〖38〗^(8 )=(3^8 ) ( 8×3^7×8^1 ) (28×3^6×8^2) (56×3^5×8^3) (70×3^4×8^4)(56×3^3×8^5) (28×3^2×8^6 ) (8×3^1×8^7) (8^8)

= (6561) (139968) (1306368) (6967296) (23224320) (49545216) (66060288) (50331648) (16777216)
= (6561) (139968) (1306368) (6967296) (23224320) (49545216) (66060288) (50331648+1677721) (6)
= (6561) (139968) (1306368) (6967296) (23224320) (49545216) (66060288) (52009369) (6)
= (6561) (139968) (1306368) (6967296) (23224320) (49545216) (66060288+5200936) (9) (6)
= (6561) (139968) (1306368) (6967296) (23224320) (49545216) (71261224) (9) (6)
= (6561) (139968) (1306368) (6967296) (23224320) (49545216+7126122) (4) (9) (6)
= (6561) (139968) (1306368) (6967296) (23224320) (56671338) (4) (9) (6)
= (6561) (139968) (1306368) (6967296) (23224320+5667133) (8) (4) (9) (6)
= (6561) (139968) (1306368) (6967296) (28891453) (8) (4) (9) (6)
= (6561) (139968) (1306368) (6967296+2889145) (3) (8) (4) (9) (6)
= (6561) (139968) (1306368) (9856441) (3) (8) (4) (9) (6)
= (6561) (139968) (1306368+985644) (1) (3) (8) (4) (9) (6)
= (6561) (139968) (2292012) (1) (3) (8) (4) (9) (6)
= (6561) (139968+229201) (2) (1) (3) (8) (4) (9) (6)
= (6561) (369169) (2) (1) (3) (8) (4) (9) (6)
= (6561+36916) (9) (2) (1) (3) (8) (4) (9) (6)
= (43477) (9) (2) (1) (3) (8) (4) (9) (6)
= 4347792138496. Hence the result.

Nth power of a number

With the help of binomial theorem we can find out the value of any power of a number.

 We know that

(x+a)^n=∑_(k=0)^n▒〖(n¦k) x^k a^(n-k) 〗

 With the help of above theorem we can calculate the value of any power of a number in following way…

〖11〗^4 =〖(1〗^4) (4 ×1^3×1^1) (6 ×1^2× 1^2) (4 × 1^1×1^3) (1^4)
 = (1) (4) (6) (4) (1)
 = 14641.

〖12〗^(4 ) = (1^(4 ) ) (4× 1^(3 )× 2^(1 ) ) (6×1^(2 )×2^(2 ) ) (4×1^(1 )×2^(3 ) )(2^(4 ) )
 = (1) (8) (24) (32) (16)
 = (1) (8) (24) (32+1) (6)
 = (1) (8) (24+3) (3) (6)
 = (1) (8+2) (7) (3) (6)
 = (1+1) (0) (7) (3) (6)
 = (2) (0) (7) (3) (6)
 = 20736.

〖21〗^(4 ) =〖(2〗^(4 )) (4×2^(3 )×1^(1 ) ) (6×2^(2 )×1^(2 ) ) (4×2^(1 )×1^(3 ) ) (1^(4 ))
= (16) (32) (24) (8) (1)
= (16) (32+2) (4) (8) (1)
= (16) (34) (4) (8) (1)
= (16+3) (4) (4) (8) (1)
= (19) (4) (4) (8) (1)
= 194481.

Transfer of property to unborn person

Who is unborn person? An unborn person means a person who is not in existence even in mother’s womb.

According to section 5 of Transfer of Property Act 1882, transfer of property takes place only between two living persons (see the language of sec 5 of the said Act-“transfer of property” means an act by which a living person conveys property, in present or in future, to one or more other living persons, or to himself, or to himself and one or more other living persons, and “to transfer of property” is to perform such act). This means that transferee i.e. the person to whom it is transferred, must also be in existence at the date of transfer.

In the light of the section 5 of the T.P.Act, property cannot be transferred directly to an unborn person.  In between the transferor and the unborn there must be an intermediary living person who may hold the property in trust for the benefit of the unborn. The unborn must come into existence before the death of the person holding property for life. If the unborn comes into existence after the death of the last living person, the property shall revert back to the transferor or his heirs.

Section 13 of T.P.Act 1882, read as “where, on a transfer of property, an interest therein is created for the benefit of a person not in existence at the date of the transfer, subject to a prior interest created by the same transfer, the interest created for the benefit of such person shall not take effect, unless it extends to the whole of the remaining interest of the transferor in the property”.

Illustration: -- A transfer his property to X for life who is unmarried and then to the eldest child of X absolutely. The transfer in favor of eldest child of X is valid.

Giving life interest or creating life-estate in favor of a person means giving him only the right of enjoyment and possession. He has to preserve the property like a trustee during his life time on behalf of the unborn. If absolute interest is given to this living person, he may be entitled to dispose it of to anyone.

 Further in the light of section 16 of the T.P. Act, if prior interest created under section 13 fails, the subsequent interest depending on it also fails.

In the case of Girish Dutt V/s Data Din, the gift to unborn daughters with no powers of alienation was held to be invalid.


Disclaimer- All the content are for general use and information. Consult your lawyer before acting

Cube: Try it

Cube: Try it

With the help of

 (a+b) ^3 = a^3 + 3a^2b +3ab^2 + b^3,

We can easily find out the cube of any number in the following manner

11^3 = (1^3) (3×1^2×1) (3×1×1^2) (1^3)

= (1) (3) (3) (1)

=1331

12^3 = (1^3) (3×1^2×2) (3×1×2^2) (2^3)

= (1) (6) (12) (8)

= (1) (6+1) (2) (8)

= 1728

13^3 = (1^3) (3×1^2×3) (3×1×3^2) (3^3)

= (1) (9) (27) (27)

= (1) (9) (27+2) (7)

= (1) (9) (29) (7)

= (1) (9+2) (9) (7)

= (1) (11) (9) (7)

= (1+1) (1) (9) (7)

=2197

14^3 = (1^3) (3×1^2×4) (3×1×4^2) (4^3)

= (1) (12) (48) (64)

= (1) (12) (48+6) (4)

= (1) (12) (54) (4)

= (1) (12+5) (4) (4)

= (1) (17) (4) (4)

= (1+1) (7) (4) (4)

= 2744

15^3 = (1^3) (3×1^2×5) (3×1×5^2) (5^3)

= (1) (15) (75) (125)

= (1) (15) (75+12) (5)

= (1) (15) (87) (5)

= (1) (15+8) (7) (5)

= (1) (23) (7) (5)

= (1+2) (3) (7) (5)

= 3375

26^3 = (2^3) (3×2^2×6) (3×2×6^2) (6^3)

= (8) (3×4×6) (3×2×36) (216)

= (8) (72) (216) (216)

= (8) (72) (216+21) (6)

= (8) (72) (237) (6)

= (8) (72+23) (7) (6)

= (8) (95) (7) (6)

= (8+9) (5) (7) (6)

= 17576

37^3 = (3^3) (3×3^2×7) (3×3×7^2) (7^3)

= (27) (189) (441) (343)

= (27) (189) (441+34) (3)

= (27) (189+47) (5) (3)

= (27+23) (6) (5) (3)

=50653

98^3 = (9^3) (3×9^2×8) (3×9×8^2) (8^3)

= (729) (1944) (1728) (512)

= (729) (1944) (1728+51) (2)

= (729) (1944+177) (9) (2)

= (729+212) (1) (9) (2)

= 941192.

Multiplication Game

Easy Multiplication

With the help of distributive law (a+b) ×(c+d) = ac + ad + bc + bd; we can easily multiply two numbers in the following way

26×34 = (2×3) (2×4) (6×3) (6×4)

= (6) (8) (18) (24)

= (6) (8 + 18) (24)

= (6) (26) (24)

= (6) (26+2) (4)

= (6) (28) (4)

= (6+2) (8) (4)

= (8) (8) (4)

= 884.

18629×96 = (1×9) (1×6) (8×9) (8×6) (6×9) (6×6) (2×9) (2×6) (9×9) (9×6)

= (9) (6) (72) (48) (54) (36) (18) (12) (81) (54)

= (9) {(6) + (72)} {(48) + (54)} {(36) + (18)} {(12) + (81)} (54)

= (9) (78) (102) (54) (93) (54)

= (9) (78) (102) (54) (93+5) (4)

= (9) (78) (102) (54) (98) (4)

= (9) (78) (102) (54+9) (8) (4)

= (9) (78) (102) (63) (8) (4)

= (9) (78) (102+6) (3) (8) (4)

= (9) (78) (108) (3) (8) (4)

= (9) (78+10) (8) (3) (8) (4)

= (9) (88) (8) (3) (8) (4)

= (9+8) (8) (8) (3) (8) (4)

= (17) (8) (8) (3) (8) (4)

= 1788384.

195264×105

 = 195264×105 = (1×10) (1×5) (9×10) (9×5) (5×10) (5×5) (2×10) (2×5) (6×10) (6×5) (4×10) (4×5)

 = (10) (5) (90) (45) (50) (25) (20) (10) (60) (30) (40) (20)

= (10) (5+90) (45+50) (25+20) (10+60) (30+40) (20)

= (10) (95) (95) (45) (70) (70) (20)

= (10) (95) (95) (45) (70) (70+2) (0)

= (10) (95) (95) (45) (70) (72) (0)

= (10) (95) (95) (45) (70+7) (2) (0)

= (10) (95) (95) (45) (77) (2) (0)

= (10) (95) (95) (45+7) (7) (2) (0)

= (10) (95) (95) (52) (7) (2) (0)

= (10) (95) (95+5) (2) (7) (2) (0)

= (10) (95) (100) (2) (7) (2) (0)

= (10) (95+10) (0) (2) (7) (2) (0)

= (10) (105) (0) (2) (7) (2) (0)

= (10+10) (5) (0) (2) (7) (2) (0)

= (20) (5) (0) (2) (7) (2) (0)

= 20502720.

Square: Some more examples

(101)^2 = (10^2) (2×10×1) (1^2) or (1^2) (2×1×01) {(01) ^2}

               = (100) (20) (1) or (1) (02) (01)

               = (100+2) (0) (1) or 10201

               = (102) (0) (1) or 10201

                = 10201 or 10201.

(102)^2 = (10^2) (2×10×2) (2^2) or (1^2) (2×1×02) {(02) ^2}

               = (100) (40) (4) or (1) (04) (04)

               = (100+4) (0) (4) or 10404

               = (104) (0) (4) or 10404

               = 10404 or 10404.

(105)^2 = (10^2) (2×10×5) (5^2) or (1^2) (2×1×05) {(05) ^2}

               = (100) (100) (25) or (1) (10) (25)

               = (100) (100+2) (5) or 11025

               = (100) (102) (5) or 11025

               = (100+10) (2) (5) or 11025

               = (110) (2) (5) or 11025

               = 11025 or 11025.

(124)^2 = (12^2) (2×12×4) (4^2) or (1^2) (2×1×24) (24^2)

               = (144) (96) (16) or (1) (48) (576)

               = (144) (96+1) (6) or (1) (48+5) (76)

               = (144) (97) (6) or (1) (53) (76)

               = (144+9) (7) (6) or 15376

               = (153) (7) (6) or 15376

                = 15376 or 15376.

(1835)^2 = (18^2) (2×18×35) (35^2)

                 = (324) (1260) (1225)

                 = (324) (1260+12) (25)

                 = (324) (1272) (25)

                 = (324+12) (72) (25)

                 = (336) (72) (25)

                 = 3367225.

(1836)^2 = (18^2) (2×18×36) (36^2)

                 = (324) (1296) (1296)

                 = (324) (1296+12) (96)

                 = (324) (1308) (96)

                 = (324+13) (08) (96)

                 = (337) (08) (96)

                 = 3370896.

Square: Try it

Square: Try it We know that
(a+b)^2 = a^2 + 2ab + b^2.
With the help of above we can find out the square of a number in the following way…
(11)^2 = (1^2) (2×1×1) (1^2)
= (1) (2) (1)
=121.
(12)^2 = (1^2) (2×1×2) (2^2)
= (1) (4) (4)
= 144.
(13)^2 = (1^2) (2×1×3) (3^2)
= (1) (6) (9)
= 169.
(14)^2 = (1^2) (2 ×1×4) (4^2)
= (1) (8) (16)
= (1) (8+1) (6)
= (1) (9) (6)
= 196.
(15)^2 = (1^2) (2 ×1×5) (5^2)
= (1) (10) (25)
= (1+1) (0+2) (5)
= (2) (2) (5)
=225.
(16)^2 = (1^2) (2×1×6) (6^2)
= (1) (12) (36)
= (1+1) (2+3) (6)
= (2) (5) (6)
= 256.
(17)^2 = (1^2) (2×1×7) (7^2)
= (1) (14) (49)
= (1+1) (4+4) (9)
= (2) (8) (9)
= 289.
(18)^2 = (1^2) (2×1×8) (8^2)
= (1) (16) (64)
= (1+1) (6+6) (4)
= (2) (12) (4)
= (2+1) (2) (4)
= (3) (2) (4)
= 324
(19)^2 = (1^2) (2×1×9) (9^2)
= (1) (18) (81)
= (1+1) (8+8) (1)
= (2) (16) (1)
= (2+1) (6) (1)
= 361.
(20)^2 = (2^2) (2×2×0) (0^2)
= (4) (0) (0)
= 400.
(21)^2 = (2^2) (2×2×1) (1^2)
= (4) (4) (1)
= 441.
(25)^2 = (2^2) (2×2×5) (5^2)
= (4) (20) (25)
= (4+2) (0+2) (5)
= (6) (2) (5)
= 625.
(98)^2 = (9^2) (2×9×8) (8^2)
= (81) (144) (64)
= (81) (144+6) (4)
= (81) (150) (4)
= (81+15) (0) (4)
= (96) (0) (4)
= 9604.

square of a natural number is the sum of two definite natural numbers.

Square of a natural number can be found with the help of the following equation

(xy)^2 = x(x+1)y^2 + 20x(y−5).

Where y = 0 to 9; x = 0 to ∞.

Explanation:- value of first number is obtained by putting the value of x(x+1) next to the value of y^2.

Example:- square of 37 can be obtained by putting the value of 3(3+1) next to the value of 7^2 and add into them 20 times the value of 3(7−5).

Now 3(3+1) = 3×4 = 12 and 7^2 =49, put them next to each other. We get 1249.
In 1249 add 20 times the value of 3×(7−5).

Now 20×3(7−5)= 20×3×2 = 120

So 1249+120=1369. This is the required square of 37.

Another example
596^2 = 59(59+1)6^2 +20×59(6−5).
=(3540)36 + 1180
=354036+1180
=355216. This is the required square of 596. Here value of y is 6 and value of x is 59.

Another example
594^2 = 59(59+1)4^2 + 20×59(4−5)
= (59×60)4^2 + 20×59×(−1)
= (3540)16 − 1180
=354016 – 1180
=352836. This is the required square of 594. Here value of y is 4 and value of x is 59.

Note:- Numbers has its mirror image.