Solve:(1-1/2²)(1-1/3²)(1-1/4²)... ... ....(1-1/10²)=??
Answers
Answer:
The series you mentioned is a special type of series called, hyper harmonic series.
According to me, there doesn’t exist any specific formula for finding this sum yet, but ya we are having many tests ( for example, p series test, ratio test etc.) to prove that the series converges as n tends to infinity (n->\infty).
However, I’ll prove here by a simple way, which do not require you any prior knowledge(of such tests/theorems) to find the result.
We know that :-
(1/n^2) < = [1/{n*(n-1)}] = 1/(n-1) - 1/n …………………………………….. eqn(1)
Let us denote sum of first n terms by Sn…
i.e. S1=1/1^2
S2=1/1^2 + 1/2^2
……………………
Sn=(1/1^2) + (1/2^2) + … + (1/n^2)
Now, from eqn(1):-
Sn <= (1)+ [(1/1–1/2)+(1/2–1/3)+(1/3–1/4)+(1/4–1/5)+…+(1/(n-1)-1/n)]
we observe that:-
the 2nd and 3rd terms can be cancelled out by basic algebra property.
similarly the 4th and 5th terms can be cancelled out.
similarly the 5th and 6th terms can be cancelled out.
similarly the 6th and 7th terms can be cancelled out….
and so on.. till the 2nd last term i.e. 1/(n-1) is cancelled out.
so, we get:-
Sn<=1+[1–1/n]
Now, clearly as n-> \infty, then, 1/n becomes 0
So, as n tends to infinity, the Sn <= 1+[1–1/n] -> 1+[1–0]=2.
Hence the series is bounded above.
Clearly it is a convergent series.
Hope it will help..
Answer:
answer '
The given equation is 2kx
2
−40x+25=0
Here, a=2k,b=−40,c=25
∴D=b
2
−4ac=(−40)
2
−4×2k×25=1600−200k
The equation will have equal roots, if D=0⇒1600−200k=0
⇒k=8
Substituting k=8 in the given equation, we get
16x
2
−40x+25=0⇒(4x−5)
2
=0⇒x=
4
5
Hence, the roots of the given equation are each equal to
4
5
hope it helps
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