mathematical induction

Solution

Let P (n) be the proposition 3 + 2．3^2 + 3．3^3 + ．．．+ n．3^n = 3/4[1 + (2n - 1)3^n]
, where n is a natural number

(i) When n = 1
L H S =1 (3) ^1 = 1(3) = 3
R H S = (3/4) {1 + [2(1) – 1]3^1} = (3/4) [1 + (1)3^1] = (3/4) [1 + 3] = (3/4) (4) = 3
L H S = R H S
∴P (1) is true.

(ii) Assume P (k) is true, where k is a natural number
That is, 3 + 2．3^2 + 3．3^3 + ．．．+ k．3^k = 3/4[1 + (2k - 1)3^k]

L H S of P (k+1)
= 3 + 2．3^2 + 3．3^3 + ．．．+ k．3^k + (k + 1)．3^ (k + 1)
= 3/4[1 + (2k - 1)3^k] + (k + 1)．3^ (k + 1)
= 3/4[1 + (2k - 1)3^k] + 4/4 [(k + 1)．3^ (k + 1)]
= 3/4[1 + (2k - 1)3^k] + 4/4 [(k + 1)．3^ (k + 1)]
= 1/4[3 + (3) (2k - 1)3^k + 4 (k + 1)．3^ (k + 1)]
= 1/4[3 + (2k - 1)3^ (k + 1) + (4k + 4)．3^ (k + 1)]
= 1/4[3 + (2k – 1 + 4k + 4)3^ (k + 1)]
= 1/4[3 + (6k + 3)3^ (k + 1)]
= 1/4[3 + 3(2k + 1)3^ (k + 1)]
= 1/4[3 + 3(2k + 2 - 1)3^ (k + 1)]
= 3/4{1 + [2(k + 1) – 1] 3^ (k + 1)}
= R H S of P (k+1)
∴ P (k + 1) is true if P (k) is true

Combine (i) and (ii), by the principle of Mathematical Induction, P (n) is true for all natural numbers n.

Let P(n) be 3 + 2×3^2 + 3×3^3 + ．．．+ nx3^n = 3/4[ 1 + (2n - 1)3^n ] for all positive integers n.

When n = 1,

Left side
= 1(3^1)
= 3
Right side
= 3/4[1+(2(1)-1)(3^1)]
= 3/4[1+(2-1)(3)]
= 3/4[1+3]
= 3
As left side = right side,
So P(1) is true.

Assume P(k) is true, i.e. 3 + 2×3^2 + 3×3^3 + ．．．+ k.3^k = 3/4[ 1 + (2k - 1)3^k ]

When n = k+1,

Left side
= 3 + 2×3^2 + 3×3^3 + ．．．+ k×3^k + (k+1)×3^(k+1)

= (3 + 2×3^2 + 3×3^3 + ．．．+ k×3^k) + (k+1)3^(k+1) ............ by assumsion of P(k)

= 3/4[ 1 + (2k - 1)3^k ] + (k+1)3^(k+1)

= 3/4[ 1 + (2k - 1)3^k + (4/3)(k+1)3^(k+1)]

= 3/4[ 1 + (2k - 1)3^k + 4(k+1)3^(k)]

= 3/4[ 1 + (3^k)(2k - 1 + 4(k+1)]

= 3/4[ 1 + (3^k)(2k - 1 + 4k + 4)]

= 3/4[ 1 + (3^k)(6k + 3)]

= 3/4[ 1 + (3^(k+1))(2k+1)]

= 3/4[ 1 + (2(k+1)-1)3^(k+1)]

= Right side

So P(k+1) is true.

By Mathematical Induction, 3 + 2×3^2 + 3×3^3 + ．．．+ nx3^n = 3/4[ 1 + (2n - 1)3^n ] is true for all positive integers n.