Question

the great Answer of this question.......................................!


A, which is equal to 1–1+1–1+1–1 repeated an infinite number of times. I’ll write it as such:
A = 1–1+1–1+1–1⋯
Then I do a neat little trick. I take away A from 1
1-A=1-(1–1+1–1+1–1⋯)

1-A=1–1+1–1+1–1+1⋯
1-A =A
1-A+A=A+A
1 = 2A
1/2 = A

We start the same way as above, letting the series B =1–2+3–4+5–6.....This time, instead of subtracting B from 1, we are going to subtract it from A. Mathematically, we get this:
A-B = (1–1+1–1+1–1⋯) — (1–2+3–4+5–6⋯)
A-B = (1–1+1–1+1–1⋯) — 1+2–3+4–5+6⋯
A-B = (1–1) + (–1+2) +(1–3) + (–1+4) + (1–5) + (–1+6)⋯
A-B = 0+1–2+3–4+5⋯
Once again, we get the series we started off with, and from before, we know that A = 1/2,
A-B = B
A = 2B
1/2 = 2B
1/4 = B
And voila!
Once again we start by letting the series C = 1+2+3+4+5+6⋯, .
B-C = (1–2+3–4+5–6⋯)-(1+2+3+4+5+6⋯)

Because math is still awesome

B-C = (1-2+3-4+5-6⋯)-1-2-3-4-5-6⋯
B-C = (1-1) + (-2-2) + (3-3) + (-4-4) + (5-5) + (-6-6) ⋯
B-C = 0-4+0-8+0-12⋯
B-C = -4-8-12⋯
B-C = -4(1+2+3)⋯
B-C = -4C
B = -3C
And since we have a value for B=1/4, we simply put that value in and we get our magical result:
1/4 = -3C
1/-12 = C or C = -1/12

Answer 1

Answer:

oh wow

aap ne sabit kar diya ki aap maths ke Aryabhatta hai

congratulations

Step-by-step explanation:

agar koi maths Mai doubt hua to aapse puch sakti hu kya

Answer 2

Answer:

Speed (v) abs wavelength ( �) and the frequency (v) of sound are related as

a. λ= v X v

b. v= λ X v

c. v= λ X v

d. v= λ / v

solve it fast Mr penguin.............