Find the sum upto 2n+1 terms :-
2+3+5+9+8+15+11+21+...
Answers
Step-by-step explanation:
Given series is:
S = {2 + 3 + 5 + 9 + 8 + 15 + 11 + 21 + ----} to 2n + 1 terms
The given series can be rearranged as –
S = {2 + 5 + 8 + 11 + 14 + ----} + {3 + 9 + 15 + 21 + ----}
Since the total number of terms in the series are odd (i.e. 2n + 1), so series in first bracket will have n + 1 terms and series in second bracket will have n terms.
Let S1 = {2 + 5 + 8 + 11 + 14 + ---- upto n +1 terms}
and S2 = {3 + 9 + 15 + 21 + ---- upto n terms}
∴ S = S1 + S2 .......... (1)
firstly take, S1 = 2 + 5 + 8 + 11 + 14 + ---- upto n +1 terms
Since a = 2 , d = 5 – 2 = 3 , N (number of terms) = n +1
∴ l = a + (N – 1) d [N is total number of terms]
⇒ l = 2 + (n + 1 – 1) × 3
⇒ l = 2 + 3n
Consider the second series
S2 = 3 + 9 + 15 + 21 + .... n terms
Here a = 3, d = 9 – 3 = 6 and number of terms (N) = n
From (1) and (2), Sum of given series is –
S2n + 1 = (Sum of S1) + (Sum of S2)
(some calculations are in the above attachment )
ANSWER :
Explanation:
S = {2+3+5+9+8+15+11+21+.......} to (2n+1) terms.
Let the series be divided into two groups.
S1= {2+5+8+11+14+.....}
S2= {3+9+15+21+......}
Since, the no of terms (i.e. 2n+1) signifies odd no of terms
Therefore, the no of terms in S1 is (n+1) terms.
Consider the example,
a+b+c+d+e+f+g+h+i+j+k (11 terms) which is the form of 2n+1 where n=5
If the series is rearranged to be a+c+e+g+i+k then the number of terms is 6 (ie. no of terms is 2n+1+1/2)
therefore the rest series includes b + d + f + h + j which is 5 terms (that is number of terms is n)
Therefore,
Using formula, Sn=n/2[2a+(n-1)d] where 'n' is the number of terms, 'a' is the first term, 'd' is the common difference.
In case of S1, a=2, d=5-2=3
In case of S2, a=3, d=9-3=6.
Now, Sn=S1+S2
Simplifying,