Tamil Nadu Board of Secondary EducationHSC Arts Class 11th

Find the sum up to the 17th term of the series 131+13+231+3+13+23+331+3+5+... - Mathematics

Advertisement Remove all ads
Advertisement Remove all ads
Advertisement Remove all ads
Sum

Find the sum up to the 17th term of the series `1^3/1 + (1^3 + 2^3)/(1 + 3) + (1^3 + 2^3 + 3^3)/(1 + 3 + 5) + ...` 

Advertisement Remove all ads

Solution

t1 = `1^3/1`

t3 = `(1^3 + 2^3)/(1 + 3)`

t3 = `(1^3 + 2^3 + 3^3)/(1 + 3 + 5)`

∴ tn = `(1^3 + 2^3 + 3^3 + ... :n:^3)/(1 + 3 + 5 + ... "n terms")`

= `(sum"n"^3)/("n"^2) ((2"n" - 1 + 1)/2)^2`

= `("n"^2("n" + 1)^2)/(4("n"^2))`

= `("n" + 1)^2/4`

= `("n"^2 + 2"n" + 1)/4`

∴ Sn = `sum"t"_"n" = 1/4 sum"n"^2 + 2"n" + 1`

= `1/4 sum"n"^2 + sum2"n" + sum1`

= `1/4[("n"("n" + 1)(2"n" + 1))/6 + (2("n")("n" + 1))/2 + "n"]`

To find S17 put n = 17

S17 = `1/4[(17 xx 18 xx 35)/6 + (2(17)(18))/2 + 17]`

= `1/4[17 xx 105 + 17 xx 18 + 17]`

= `(17(105 + 18 + 1))/4`

= 17 × 31

= 527

Concept: Finite Series
  Is there an error in this question or solution?

APPEARS IN

Tamil Nadu Board Samacheer Kalvi Class 11th Mathematics Volume 1 and 2 Answers Guide
Chapter 5 Binomial Theorem, Sequences and Series
Exercise 5.3 | Q 2 | Page 220
Share
Notifications

View all notifications


      Forgot password?
View in app×