(i) \(\displaystyle\sum_{n=1}^{11} (2 + 3^n)\)
= (2 + 31) + (2 + 32) + (2 + 33) + … + (2 + 311)
= 2×11 + 31 + 32 + 33 + … + 311
= 22 + 3(311 – 1)/(3 – 1) [by using the formula, a(1 – rn )/(1 – r)]
= 22 + 3(311 – 1)/2
= [44 + 3(177147 – 1)]/2
= [44 + 3(177146)]/2
= 265741
(ii) \(\displaystyle\sum_{k=1}^{n} (2^k + 3^{k - 1})\)
= (2 + 30) + (22 + 3) + (23 + 32) + … + (2n + 3n-1)
= (2 + 22 + 23 + … + 2n) + (30 + 31 + 32 + …. + 3n-1)
Firstly let us consider,
(2 + 22 + 23 + … + 2n)
Where, a = 2, r = 22/2 = 4/2 = 2, n = n
By using the formula,
Sum of GP for n terms = a(rn – 1 )/(r – 1)
= 2 (2n – 1)/(2 – 1)
= 2 (2n – 1)
Now, let us consider
(30 + 31 + 32 + …. + 3n)
Where, a = 30 = 1, r = 3/1 = 3, n = n
By using the formula,
Sum of GP for n terms = a(rn – 1 )/(r – 1)
= 1 (3n – 1)/ (3 – 1)
= (3n – 1)/2
So,
\(\displaystyle\sum_{k=1}^{n} (2^k + 3^{k - 1})\)
= (2 + 22 + 23 + … + 2n) + (30 + 31 + 32 + …. + 3n)
= 2 (2n – 1) + (3n – 1)/2
= 1/2 [2n+2 + 3n – 4 – 1]
= 1/2 [2n+2 + 3n – 5]
(iii) \(\displaystyle\sum_{n=2}^{10} 4^n\)
= 42 + 43 + 44 + … + 410
Where, a = 42 = 16, r = 43/42 = 4, n = 9
By using the formula,
Sum of GP for n terms = a(rn – 1 )/(r – 1)
= 16 (49 – 1)/(4 – 1)
= 16 (49 – 1)/3
= 16/3 [49 – 1]