Question

prove the following equation.​

Answer 1

Step-by-step explanation:

(a).1–1+1–1+1–1 ⋯

(b).1–2+3–4+5–6⋯

Let 1–1+1–1+1–1 ⋯=A

A=1–1+1–1+1–1 ⋯

1+A=1+(1-1+1-1⋯)

multiple by - on both side

1-A=1-(1–1+1–1+1–1⋯)

1-A=1–1+1–1+1–1+1⋯

1-A =A

1/2 = A

Let B=1–2+3–4+5–6⋯

consider A-B

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⋯

A-B = B

A = 2B

1/2 = 2B

1/4 = B

NOW LET C=1+2+3+4+5+6⋯

consider B-C

B-C = (1–2+3–4+5–6⋯)-(1+2+3+4+5+6⋯)

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

1/4 = -3C

1/-12 = C or C = -1/12

Answer 2

$\underline{\underline{\bf{Question}}}$

Prove that :-

$1 + 2 + 3 + 4 + 5 + .... \infty = \dfrac{- 1}{12}$

$\rule{190pts}{2pts}$

Solution

$\sum\limits_{ \sf \: n=1}^{ \infty } = \dfrac{-1}{12}$

Assume that ,

$\sf \: s_1 = (1 -1) + (1 - 1) + (1 - 1) + (1 - 1).... \sf \: even \\ \therefore \boxed{ \sf \: s_1 = 0} \\ \tt \: s_1 = 1( - 1 + 1)( - 1 + 1)( - 1 + 1)( - 1 + 1)... \sf odd \\ \therefore\boxed{ \sf \: s_1 = 1}$

$\sf \: s_1 = \dfrac{0 + 1}{2} = \dfrac{1}{2} \rightarrow(1)$

$\sf \: s_2 = 1 - 2 + 3 - 4 + 5 - 6 + 7... \infty \\ \sf s_2 = 1 - 2 + 3 - 4 + 5 - 6 + 7... \infty \\ \sf \: 2s_1 = 1 - 1 + 1 - 1 + 1 - 1 = 0 \\ \sf 2s_2 = s_1 = \dfrac{1}{2} \\ \sf s_1 = \dfrac{1}{4} \rightarrow \: (2)$

Here , we will now substitute the above equation .

$\sf \: s_3 = 1 + 2 + 3 + 4 + 5 + 6 + 7... \infty \\ \sf s_2 = 1 - 2 + 3 - 4 + 5 - 6 + 7... \infty \\ \sf \: s_3 - s_2 = 4 + 8 + 12 + 16 + 20... \infty \\ \sf \: s_3 - s_2 = 4(1 + 2 + 3 + 4 + 5... \infty ) \\ \sf \: s_3 - s_2 = 4s_3 \\ \rm \: s_3 - \dfrac{1}{4} = 4s_3 \\ \sf3 s_3 = \frac{-1}{4} \\ \boxed{ \sf s_3 = \dfrac{ - 1}{12} }$

Hence , Proved ✔

$\rule{190pts}{2pts}$

Note

A series is the summation of a sequence. It is given by

$\sf \: S_n = a_1 + a_2 + a_3 + a_4 +……. + a_n$

In a finite series, a finite number of terms are present, whereas an infinite series consists of infinite terms.

Important Formula

1. The nth term of the arithmetic sequence or arithmetic progression (A.P) is given by. a + (n – 1) d .
2. The arithmetic mean [A.M] between a and b is [a + b] / 2
3. The nth term an of the geometric sequence or geometric progression [G.P] = a * rn–1
4. The geometric mean between a and b is G.M=$\pm \sqrt{ab}$

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