Solve

\frac{9}{1!}+\frac{16}{2!}+\frac{27}{3!}+\frac{42}{4!}+...

11 Answers

1
student ·

1
The Enlightened One - jsg ·

RPF..you are a real genius yaar..[12]

1
aposlil ·

Mast hai yaar!

1
student ·

thanks [4]

i am not at all modest [4]

1
sowmya ·

Goodness me,you are indeed a genius.

A small help though, how did you derive this
'2n^2+n+6'

Just by inspection ? or is there any shortcut for that ?

1
aposlil ·

I guess inspection...coz thats how I reached there too..

1
sowmya ·

16-9=7
27-16=11
42-27=15

11-7=4
15-11=4

after that ?

1
student ·

there is a theory to determine the degree of T_n

the method

let

T1 T2 T3 T4 T5 ......... be terms of seq

let

S1 S2 S3 S4 ........be T2-T1 ,T3-T2,.........respectively

if S1 S2 S3 .are in AP

then

Tn is quadratic in n

so put

Tn=an^2+bn+c
plug the first three terms to solve a , b,and c

if S1,S2,S3..are not in AP

then again take differnce and if that is in AP
Tn is cubic in n

likewise procced and increase the dgree each time [1]

the proof is not tuff

(x)^n-(x-a)^n will be of degree x^(n-1)

now u will realise i am not a genius , damn me , why did i tell this method

1
sowmya ·

Thanks :)

1
aposlil ·

How crude my inspection method looks in front of this...:(

39
Dr.House ·

dont think like that guruji... u got the q thats god

methods become good as time passes and as u solve more and more q

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