If k = 3 , where n is an even integer, then ∑(−3)−1 3 = ?2

We know that=12−1 (1 + )3= 1 + 31 + 32 . 2+. . . . . + 33. 3…..(i)(1 − )3= 1 − 31 + 32 . 2−. . . . . + (−1)33. 3…..(ii)Substituting the value of (ii) and (i), we get(1 + )3 − (1 − )3 = 2[ 31. + 33 . 3 + 35 . 5+...] = 2[ 31 . + 33. 3 +35 . 4+...]Putting = i√3, we get (1 + √3)3 − (1 + √3)3 = 2√3 [ 31 − 3. 33+ 33. 35+....]∴ 3− 3. 3 + 33. 3 +....= 1 [(1 + √3)3 − (1 + √3)3]1 3 52√3 ⇒ 1 2√3. 23[(1 +2√3 23)− (1 +2√3 23) ]23−1⇒√3[(cos n + sin n) − (cos n + sin n)]23−1⇒√32sin n = 0 as n is an even integer.

If k = 3 , where n is an even integer, then ∑(−3)−1 3 = ?2

We know that=12−1 (1 + )3= 1 + 31 + 32 . 2+. . . . . + 33. 3…..(i)(1 − )3= 1 − 31 + 32 . 2−. . . . . + (−1)33. 3…..(ii)Substituting the value of (ii) and (i), we get(1 + )3 − (1 − )3 = 2[ 31. + 33 . 3 + 35 . 5+...] = 2[ 31 . + 33. 3 +35 . 4+...]Putting = i√3, we get (1 + √3)3 − (1 + √3)3 = 2√3 [ 31 − 3. 33+ 33. 35+....]∴ 3− 3. 3 + 33. 3 +....= 1 [(1 + √3)3 − (1 + √3)3]1 3 52√3 ⇒ 1 2√3. 23[(1 +2√3 23)− (1 +2√3 23) ]23−1⇒√3[(cos n + sin n) − (cos n + sin n)]23−1⇒√32sin n = 0 as n is an even integer.

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If k = 3 , where n is an even integer, then ∑𝑘
(−3)𝑟−1 3𝑛𝐶 = ?
2

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Detailed Solution

We know that

𝑟=1

2𝑟−1 (1 + 𝑥)3𝑛= 1 + 3𝑛𝐶1 𝑥 + 3𝑛𝐶2 . 𝑥2+. . . . . + 3𝑛𝐶3. 𝑥3𝑛…..(i)
(1 − 𝑥)3𝑛= 1 − 3𝑛𝐶1 𝑥 + 3𝑛𝐶2 . 𝑥2−. . . . . + (−1)3𝑛𝐶3. 𝑥3𝑛…..(ii)
Substituting the value of (ii) and (i), we get
(1 + 𝑥)3𝑛 − (1 − 𝑥)𝑟3𝑛 = 2[ 3𝑛𝐶1. 𝑥 + 3𝑛𝐶3 . 𝑥3𝑛 + 3𝑛𝐶5 . 𝑥5+...] = 2[ 3𝑛𝐶1 . 𝑥 + 3𝑛𝐶3. 𝑥3𝑛 +

3𝑛𝐶5 . 𝑥4+...]
Putting 𝑥 = i√3, we get

(1 + 𝑖√3)3𝑛 − (1 + 𝑖√3)3𝑛 = 2𝑖√3 [ 3𝑛𝐶1 − 3. 3𝑛𝐶3+ 33. 3𝑛𝐶5+....]
∴ 3𝑛𝐶

− 3. 3𝑛𝐶 + 33. 3𝑛𝐶 +....= 1

[(1 + 𝑖√3)3𝑛 − (1 + 𝑖√3)3𝑛]
1 3 5

2𝑖√3
⇒ 1 2𝑖√3

. 23𝑛

[(1 +
2

𝑖√3 2

3𝑛
)

− (1 +
2

𝑖√3 2

3𝑛
) ]
23𝑛−1
⇒
𝑖√3

[(cos n𝛱 + 𝑖sin n𝛱) − (cos n𝛱 + 𝑖sin n𝛱)]
23𝑛−1
⇒
𝑖√3

2𝑖sin n𝛱 = 0 as n is an even integer.

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