I'm stuck...need help in Mathematical Induction

[Johnny Blaze]

Prove by the method of mathematical induction

3.4 + 5.7 + 7.10 + 9.13 +...+ (2n + 1)(3n + 1) = n/2 (4n^2 + 11n + 9)

*For n = 1

(2n + 1)(3n + 1) = n/2 (4n^2 + 11n + 9)
(2.1 + 1)(3.1 + 1) = 1/2 (4(1) + 11 + 9)
12 = 12 (proved)


*For n = k

3.4 + 5.7 + 7.10 + 9.13 +...+ (2k + 1)(3k + 1) = k/2 (4k^2 + 11k + 9)

*For n = k + 1

3.4 + 5.7 + 7.10 + 9.13 +...+ (2k + 1)(3k + 1) + (2(k + 1) + 1)(3(k + 1) + 1) = (k + 2)/2 (4(k + 1)^2 + 11(k + 1) + 9)

RHS:

(k + 2)/2 (4(k + 1)^2 + 11(k + 1) + 9)

Here's my working:

3.4 + 5.7 + 7.10 + 9.13 +...+ (2k + 1)(3k + 1) + (2(k + 1) + 1)(3(k + 1) + 1) = (k + 2)/2 (4(k + 1)^2 + 11(k + 1) + 9)

Since 3.4 + 5.7 + 7.10 + 9.13 +...+ (2k + 1)(3k + 1) = k/2 (4k^2 + 11k + 9), so

k/2 (4k^2 + 11k + 9) + (2k + 3)(3k + 4) = (k + 2)/2 (4(k + 1)^2 + 11(k + 1) + 9)

LHS:

k/2 (4k^2 + 11k + 9) + (2k + 3)(3k + 4)

k/2 (4k^2 + 11k + 9) + (6k^2 + 17k + 12)

k/2 (4k^2 + 11k + 9) + k/2(12k + 34 + 24/k)

k/2 (4k^2 + 11k + 9 + 12k + 34 + 24/k)

k/2 (4k^2 + 23k + 43 + 24/k)

Now I'm stucked. Can't solve LHS = RHS.
Please help me

 

[Steven G]

Prove by the method of mathematical induction

3.4 + 5.7 + 7.10 + 9.13 +...+ (2n + 1)(3n + 1) = n/2 (4n^2 + 11n + 9)

*For n = 1

(2n + 1)(3n + 1) = n/2 (4n^2 + 11n + 9)
(2.1 + 1)(3.1 + 1) = 1/2 (4(1) + 11 + 9)
12 = 12 (proved)


*For n = k

3.4 + 5.7 + 7.10 + 9.13 +...+ (2k + 1)(3k + 1) = k/2 (4k^2 + 11k + 9)

*For n = k + 1

3.4 + 5.7 + 7.10 + 9.13 +...+ (2k + 1)(3k + 1) + (2(k + 1) + 1)(3(k + 1) + 1) = (k + 2)/2 (4(k + 1)^2 + 11(k + 1) + 9)

RHS:

(k + 2)/2 (4(k + 1)^2 + 11(k + 1) + 9)

Here's my working:

3.4 + 5.7 + 7.10 + 9.13 +...+ (2k + 1)(3k + 1) + (2(k + 1) + 1)(3(k + 1) + 1) = (k + 2)/2 (4(k + 1)^2 + 11(k + 1) + 9)

Since 3.4 + 5.7 + 7.10 + 9.13 +...+ (2k + 1)(3k + 1) = k/2 (4k^2 + 11k + 9), so

k/2 (4k^2 + 11k + 9) + (2k + 3)(3k + 4) = (k + 2)/2 (4(k + 1)^2 + 11(k + 1) + 9)

LHS:

k/2 (4k^2 + 11k + 9) + (2k + 3)(3k + 4)

k/2 (4k^2 + 11k + 9) + (6k^2 + 17k + 12)

k/2 (4k^2 + 11k + 9) + k/2(12k + 34 + 24/k)

k/2 (4k^2 + 11k + 9 + 12k + 34 + 24/k)

k/2 (4k^2 + 23k + 43 + 24/k)

Now I'm stucked. Can't solve LHS = RHS.
Please help me

First do NOT use a decimal point for a times symbol.
2nd, you have an early error which is bold above

 

[MarkFL]

I would write the induction hypothesis \(P_k\) as:

[MATH]\sum_{i=1}^k\left((2i+1)(3i+1)\right)=\frac{k}{2}(4k^2+11k+9)[/MATH]
I would use the same induction step you did:

[MATH]\sum_{i=1}^k\left((2i+1)(3i+1)\right)+(2(k+1)+1)(3(k+1)+1)=\frac{k}{2}(4k^2+11k+9)+(2(k+1)+1)(3(k+1)+1)[/MATH]
[MATH]\sum_{i=1}^{k+1}\left((2i+1)(3i+1)\right)=\frac{k}{2}(4k^2+11k+9)+(2k+3)(3k+4)[/MATH]
At this point, I would compute the difference:

[MATH]\frac{k+1}{2}(4(k+1)^2+11(k+1)+9)-\frac{k}{2}(4k^2+11k+9)=(2k+3)(3k+4)[/MATH]
Hence:

[MATH]\sum_{i=1}^{k+1}\left((2i+1)(3i+1)\right)=\frac{k+1}{2}(4(k+1)^2+11(k+1)+9)[/MATH]
And so, having derived \(P_{k+1}\) from \(P_k\), we have completed the proof by induction.