Apa yang dipermudahkan? Terima kasih untuk jawapan.

Jawapan:

# 3 / sqrt10-1 / sqrt2 #.

Penjelasan:

Kami akan mencuba lebih banyak perkara berikut Penyelesaian Am:

(1-sqrt (1 ^ 2-1)) / sqrt (1 * 2) + (2-sqrt (2 ^ 2-1)) / sqrt (2.3) + (3-sqrt (3 ^ 2-1)) / sqrt (3 * 4) + … "upto m terms" #.

Jelas sekali, Ketua # n ^ (th) # Terma, iaitu, # T_n #, diberikan oleh,

# T_n = (n-sqrt (n ^ 2-1)) / sqrt (n (n + 1)) #, # = n / sqrt (n (n + 1)) - sqrt (n ^ 2-1) / sqrt (n (n + 1)) #.

# = (sqrtn * sqrtn) / (sqrtnsqrt (n + 1)) - (sqrt (n + 1) sqrt (n-1)) / (sqrtnsqrt (n +, #rArr T_n = sqrt (n / (n + 1)) - sqrt ((n-1) / n) ……………………. … (ast) #.

# "Oleh itu, jumlah" S_m = sum_ (n = 1) ^ (n = m) T_n #, # = T_1 + T_2 + T_3 + … + T_ (m-1) + T_m #, # = {cancelsqrt (1/2) -sqrt (0/1)} + {cancelsqrt (2/3) -cancelsqrt (1/2)} + {cancelsqrt (3/4) -cancelsqrt (2/3)} + … + {cancelsqrt ((m-1) / m) -cancelsqrt ((m-2) / (m-1))} + {sqrt (m / (m + 1)) - cancelsqrt ((m-1) / m)} … kerana, (ast) #,

#rArr S_m = sqrt (m / (m + 1)) - sqrt (0/1) = sqrt (m / (m + 1)) #.

Kini, menangani kes kami, # "The Reqd Sum =" S_9-T_1 = sqrt (9/10) - (1-sqrt (1 ^ 2-1)) / sqrt (1 * 2), # = sqrt (9/10) -1 / sqrt2 = 3 / sqrt10-1 / sqrt2 #.

Nikmati Matematik, dan Menyebarkan Keseronokan!