Question -
Answer -
= (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
= (2 + 30)+ (22 + 3) + (23 + 32) + … + (2n +3n-1)
= (2 + 22 +23 + … + 2n) + (30 + 31 +32 + …. + 3n-1)
Firstly let usconsider,
(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,
= (2 + 22 +23 + … + 2n) + (30 + 31 +32 + …. + 3n)
= 2 (2n –1) + (3n – 1)/2
= ½ [2n+2 +3n – 4 – 1]
= ½ [2n+2 +3n – 5]
= 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]