Istilah r _ ("th") bagi siri geometri adalah (2r + 1) cdot 2 ^ r. Jumlah n terma pertama siri adalah apa?

Jawapan:

# (4n-2) * 2 ^ n + 3 #

Penjelasan:

#S = sum_ {r = 0} ^ n 2r * 2 ^ r + sum_ {r = 0} ^ n 2 ^ r #

# S = sum_ {r = 1} ^ n r * 2 ^ (r + 1) + (1 - 2 ^ {n + 1}) / (1 - 2) #

# S = a_ {01} (1 - 2 ^ n) / (1- 2) + … + a_ {0n} (1 - 2 ^ {n- (n-1)}) / (1- 2) + 2 ^ {n + 1} - 1 #

#1*2^2 + 1*2^3 + 1*2^4#

#+ 1 * 2^3 + 1 * 2^4#

#+ 1 * 2^4#

#S = sum_ {i = 0} ^ {n-1} 2 ^ {i + 2} (2 ^ (n - i) - 1)

#S = 4 sum_ {i = 0} ^ {n-1} (2 ^ n - 2 ^ i) + 2 ^ {n + 1} - 1 #

#S = 4 * 2 ^ n * n - 4 * (2 ^ n - 1) + 2 ^ {n + 1} - 1 #

#S = (4n-2) * 2 ^ n + 3 #

Mari sahkan

#S = 1 * 2 ^ 0 + 3 * 2 ^ 1 + 5 * 2 ^ 2 + 7 * 2 ^ 3 + cdots #

#S = 1 + 6 + 20 + 56 + cdots #

#S (0) = 1 = -2 + 3 #

#S (1) = 7 = 2 * 2 + 3 #

#S (2) = 27 = 6 * 2 ^ 2 + 3 #

Dan #S (3) = 83 = 10 * 2 ^ 3 + 3 #

Jawapan:

# (4n-2) 2 ^ n + 2, atau, (2n-1) 2 ^ (n + 1) + 2 #

Penjelasan:

Biarkan # S_n # menunjukkan jumlah yang pertama # n # terma urutan

# (2r + 1) 2 ^ r #.

Kemudian, # S_n = sum_ (r = 1) ^ (r = n) (2r + 1) 2 ^ r #,

(2n + 1) * 2 ^ (n-1) + (2n + 1) * 2 ^ n #.

Mengalikan oleh #2#, kita mendapatkan, # 2S_n = 3 * 2 ^ 2 + 5 * 2 ^ 3 + … + (2n-1) * 2 ^ n + (2n + 1) * 2 ^ (n + 1).

(2n + 1)} 2 ^ n + (2n + 1) # 2S_n-S_n = (3-5) 2 ^ 2 ^ (n + 1) warna (merah) (- 3 * 2 ^ 1) #.

#:. S_n = warna (merah) (-2 * 2 ^ 1) -2 * 2 ^ 2-2 * 2 ^ 3-2 * 2 ^ n + (2n + 1) 2 ^ (n + 1)) (- 1 * 2 ^ 1), #

2 = 2 ^ 2 + 2 ^ 3 + … + 2 ^ n + (2n + 1) warna (biru) (2 ^ (n + 1)) - 2 #,

# 2 - 2 {2 (2 ^ n-1)} / (2-1) + (2n + 1) warna (biru) (2 ^ n * 2) -2 #, # = - 4 * 2 ^ n + 4 + (4n + 2) 2 ^ n-2 #.

# = 2 ^ n {-4+ (4n + 2)} + 4-2 #.

# rArr S_n = (4n-2) 2 ^ n + 2, atau, S_n = (2n-1) 2 ^ (n + 1) + 2 #