Question

Pretty hard Algebra!!

Prove that:

Answer 1

\[\frac{ 1 }{30 }k(k+1)(2k+1)(3k^2+3k-1+(k+1)^4 = \frac{ 1 }{ 30 }(k+1)[(k+1)+1][2(k+1)+1][3(k+1)^2+3(k+1)-1]\]

Answer 2

We have to simplify L.H.S

Answer 3

@koli123able
correct the closing parenthesis

Answer 4

This L.H.S\[\frac{ 1 }{30 }k(k+1)(2k+1)(3k^2+3k-1)+(k+1)^4 \]

Answer 5

and make it \[\frac{ 1 }{ 30 }(k+1)[(k+1)+1][2(k+1)+1][3(k+1)^2+3(k+1)-1]\]

Answer 6

i can't do it

Answer 7

substitute k=k+1 into LHS

Answer 8

@koli123able
i think u r doing some mathematical Induction problem

Answer 9

actually no , that wudnt work sorry

Answer 10

yes this topic is related to mathematical induction

Answer 11

true.... this looks like proof by induction

Answer 12

i was just asking the thing where i got stuck

Answer 13

The real question is 1^4+2^4+3^4+....+n^4= 1/30 n(n+1) (2n+1)(3n^2+3n-1)

Answer 14

prove by mathematical induction

Answer 15

always give us the real question! haha

Answer 16

(1) Put n=1 in P(n)

1^4=1/30.1(1+1)(2+1)3+3-1)
1=1
P(n) is true for n=1

Answer 17

firs take 1/30 *(k+1) common out

Answer 18

(2)Assume that P(n) is true for n=k

Answer 19

1^4+2^4+3^4+....+k^4=1/30 and so on

Answer 20

Add next term on both sides i.e, (k+1)^4

Answer 21

\[\frac{ 1 }{30 }k(k+1)(2k+1)(3k^2+3k-1)+(k+1)^4 \]

Answer 22

so this was the whole thing

Answer 23

it all comes down to the prove question i asked

Answer 24

i cant prove it

Answer 25

i have done all of the questions by same method except this one.....

Answer 26

u have alot of expnding to do and u have to use synthetic division... i hope this helps