Ex 9.4.4

Transcript

Ex 9.4, 4 Find the sum to n terms of the series 1/(1 × 2) + 1/(2 × 3) + 1/(3 × 4) +……… nth term of series 1/(1 × 2) + 1/(2 × 3) + 1/(3 × 4) +……… is 1/(𝑛(𝑛 + 1)) ⇒ an = 1/(𝑛(𝑛 + 1)) = ((𝑛 + 1) − 𝑛)/(𝑛(𝑛 + 1)) = ((𝑛 + 1))/(𝑛(𝑛 + 1)) – 𝑛/(𝑛(𝑛 + 1)) = 1/𝑛 – 1/((𝑛 + 1)) Now, an = 1/𝑛 – 1/((𝑛 + 1)) Adding all terms Sn = ("1 " −" " 1/2) + (1/2 " " −" " 1/3) + (1/3 " " −" " 1/4) + … +(1/(𝑛 − 1) " " −" " 1/𝑛)+ (1/𝑛 " " −" " 1/(𝑛 + 1)) Sn = 1 − 1/2 + 1/2 − 1/3 + 1/3 − 1/4 + … + 1/(𝑛 − 1) " "−" " 1/𝑛+ 1/𝑛 − 1/(𝑛 + 1) Sn = 1 − 1/(𝑛 + 1) Sn = ((𝑛 + 1) − 1)/(𝑛 + 1) Sn = (𝑛 + 1 − 1)/(𝑛 + 1) Sn = (𝑛 + 0)/(𝑛 + 1) Sn = 𝑛/(𝑛 + 1) Thus, the required sum is 𝑛/(𝑛 + 1)