Question

Either (A) an arithmetic sequence its terms do not match and n belongs to N *
1) Prove that (1/U1 * U2) + 1 (U2 * U3) + .... +1 / (A * A +1) = n / (U1 * A +1)

Answer 1

What is N* ?

Answer 2

naturals i guess \(n \in \mathbb{N}\)

Answer 3

yes

Answer 4

i dont really understand the first sentence

Answer 5

can you speak or understand french

Answer 6

i cannot speak french sorry

Answer 7

ok the first sentence: (Un) an arithmetic sequence

Answer 8

Oh, so you mean below is the arithmetic sequence ?
U1, U2, U3 ....

Answer 9

yes exactly !

Answer 10

given that, u want to prove below ?

\(\large \frac{1}{U1 * U2} + \frac{1}{U2 * U3} + .... + \frac{1 }{A * A +1} = \frac{n }{ U1 * A +1}\)

Answer 11

n = number of terms on left series
A = ?

Answer 12

yes but not A it's (Un) okéy ?

Answer 13

now it makes some sense. so u wanto prove below :-

\(\large \frac{1}{U1 * U2} + \frac{1}{U2 * U3} + .... + \frac{1 }{U_n * U_{n+1}} = \frac{n }{ U1 * U_{n+1}}\)

?

Answer 14

Yes exactly

Answer 15

So What should I do

Answer 16

can we use induction ?

Answer 17

YeS we can

Answer 18

have you found the solution

Answer 19

?

Answer 20

To prove :- \(\large \frac{1}{U1 * U2} + \frac{1}{U2 * U3} + .... + \frac{1 }{U_n * U_{n+1}} = \frac{n }{ U1 * U_{n+1}} \)

rewriting the given statement :-
To prove :- \(\large \frac{1}{a(a+d)} + \frac{1}{(a+d)(a+2d)} + .... + \frac{1 }{(a+(n-1)d) * (a+nd)} = \frac{n }{ a(a+nd)} \)

clearly, for \(n = 1\), this is true. so, \( n=1\) is in the solution set

assume the given statement is true for \(n = k\)
\(\large \frac{1}{a(a+d)} + \frac{1}{(a+d)(a+2d)} + .... + \frac{1 }{(a+(k-1)d) * (a+kd)} = \frac{k }{ a(a+kd)} \)

Answer 21

Did you find it

Answer 22

we need to the formula works for \(n = k+1\)

Answer 23

sum of \(k+1\) terms :-

\(\large \frac{1}{a(a+d)} + \frac{1}{(a+d)(a+2d)} + .... + \frac{1 }{(a+(k-1)d) * (a+kd)} +\frac{1 }{(a+kd) * (a+(k+1)d)} \)

from the induction hypothesis, sum till k terms equals k/(a(a+kd))

\(\large \frac{k }{a * (a+kd)} +\frac{1 }{(a+kd) * (a+(k+1)d)} \)

\(\large \frac{k* (a+(k+1)d) }{a * (a+kd)* (a+(k+1)d)} +\frac{1*a }{a*(a+kd) * (a+(k+1)d)} \)

\(\large \frac{k* (a+(k+1)d) + a }{a * (a+kd)* (a+(k+1)d)} \)

\(\large \frac{ka +k(k+1)d + a }{a * (a+kd)* (a+(k+1)d)} \)

\(\large \frac{a(k+1) +k(k+1)d }{a * (a+kd)* (a+(k+1)d)} \)

\(\large \frac{(k+1)(a+kd) }{a * (a+kd)* (a+(k+1)d)} \)

\(\large \frac{(k+1)}{a (a+(k+1)d)} \)

so, \(k+1\) th term also lies in the solution set.

QED

Answer 24

see if that makes sense

Answer 25

ok