Question

Find the sum upto 2n+1 terms :-
2+3+5+9+8+15+11+21+...

Answer 1

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 2

ANSWER :

$\frac{1}{2} (9 {n}^{2} + 7n + 4)$

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

$s(n + 1) = \frac{n + 1}{2} (2 \times 2 + (n + 1- 1)3) \\ = \frac{n + 1}{2} (4 + 3n)$

In case of S2, a=3, d=9-3=6.

$s(n) = \frac{n}{2} (2 \times 3 + (n - 1)6) \\ = \frac{n}{2} (6 + 6n - 6) \\ = \frac{n}{2} \times 6n \\ = 3 {n}^{2}$

Now, Sn=S1+S2

$s(2n + 1) = s(n + 1) + s(n) \\ = \frac{n + 1}{2} (4 + 3n) + 3 {n}^{2}$

Simplifying,

$\frac{1}{2} (9 {n}^{2} + 7n + 4)$