Find the sum Sigma4k1 2k3 sigma5i0 i2SolutionThe solution is

Find the sum. Sigma^4_k=1 (2k+3) sigma^5_i=0 i^2

Solution

The solution is as under:

(a) The kth termof the series is 2k + 3 . The first term, when k = 1 is 2 + 3 = 5, The (k-1) th term is 2(k-1) + 3 = 2k -2 + 3 = 2k +1 . The difference between the kth and the (k-1)th terms is 2k +3 - 2k -1 = 2 . Thus the given series is an arithmetic series with the first term as 5 and with a common difference of 2. We know that the sum of an arithmetic series of n terms, where a is the first term and b is the common difference, is given by S = n/2 [ 2a + (n -1)b] . Here, n = 4, a = 5 and b = 2,. Therefore, the required sum is 4/2[ 2*5 + ( 4 - 1)2] = 2( 10 + 6] = 32.

(b) We know that the sum of the series 0² + 1² + 2² +...+ n² = {n (n + 1) (2n + 1)} / 6. Here, n = 5. Therefore, the required sum is 1/6 { 5 * 6*11} = 330/6 = 55