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Evaluate the following:

(i) \(\displaystyle\sum_{n=1}^{11} (2 + 3^n)\)

(ii) \(\displaystyle\sum_{k=1}^{n} (2^k + 3^{k - 1})\)

(iii) \(\displaystyle\sum_{n=2}^{10} 4^n\)

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(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]

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