Find the sum to n terms of 1/(1.2.3) + 3/(2.3.4) + 5/(3.4.5) + 7/(4.5.6)
I get
n(3n+1)
---------------
4(n^2+3n+2)
How did you do it?
Yes,your answer is correct. Can you please show your workings?
use induction. Let the sum be f(n)
for k=1, it is clear that
1/(1*2*3) = 1(4)/(4*6)
Now assume true for k=n and see what happens for k=n+1
1/(1*2*3) + ... + (2n-1)/(n(n+1)(n+2)) + (2(n+1)-1)/((n+1)(n+2)(n+3))
= n(3n+1)/(4(n+1)(n+2)) + (2n+1)/(n+1)(n+2)(n+3)
=
n(3n+1)(n+3) + 4(2n+1)
------------------------
4(n+1)(n+2)(n+3)
=
3n^3+10n^2+3n+8n+4
-----------------------
4(n+1)(n+2)(n+3)
=
3n^3+10n^2+11n+4
----------------------
4(n+1)(n+2)(n+3)
=
(n+1)^2(3n+4)
------------------
4(n+1)(n+2)(n+3)
=
(n+1)(3(n+1)+1) / 4(n+1)(n+2)
which is f(n+1)
The sum of the first "n" termsof a series 2^(n+3) -8.Find the 7th term.
To find the sum of the given series, let's first observe the pattern in the terms:
1/(1.2.3), 3/(2.3.4), 5/(3.4.5), 7/(4.5.6)
Notice that the numerator of each term is an odd number starting from 1, and the denominators form a sequence of consecutive numbers.
Now, let's break down the terms:
Term 1: 1/(1.2.3) = 1/6
Term 2: 3/(2.3.4) = 3/24 = 1/8
Term 3: 5/(3.4.5) = 5/60 = 1/12
Term 4: 7/(4.5.6) = 7/120 = 1/17.14
We can see that the denominator of each term is given by the multiplication of three consecutive numbers. Therefore, for the nth term, the denominator will be (n+1)(n+2)(n+3).
Now, let's calculate the general term of the series:
General term, T(n) = (2n-1) / [(n+1)(n+2)(n+3)]
To find the sum of n terms, we need to add up all the individual terms from 1 to n:
Sum of n terms, S(n) = T(1) + T(2) + T(3) + ... + T(n)
Substituting the value of T(n) into the sum equation, we have:
S(n) = [(2*1-1) / (2*1+1)(2*1+2)(2*1+3)] + [(2*2-1) / (2*2+1)(2*2+2)(2*2+3)] + [(2*3-1) / (2*3+1)(2*3+2)(2*3+3)] + ... + [(2*n-1) / (2*n+1)(2*n+2)(2*n+3)]
Simplifying this equation gives us the sum of the series to n terms.