IF `P=2^2+6^2+10^2+14^2+…..94^2` and Q =`1^2+5^2+9^+…….81^2` then what is the value of P-Q

Options

A. 24645
B. 26075
C. 29317
D. 31515

Answer

To solve the problem, we need to calculate the values of \( P \) and \( Q \) and then find \( P - Q \).

### Step 1: Calculate \( P \)

The series for \( P \) is given by:
\[ P = 2^2 + 6^2 + 10^2 + 14^2 + \ldots + 94^2 \]

This series consists of squares of numbers that form an arithmetic sequence where the first term \( a = 2 \) and the common difference \( d = 4 \).

To find the number of terms in this series, we can use the formula for the \( n \)-th term of an arithmetic sequence:
\[ a_n = a + (n-1)d \]
Setting \( a_n = 94 \):
\[ 94 = 2 + (n-1) \cdot 4 \]
\[ 92 = (n-1) \cdot 4 \]
\[ n-1 = 23 \]
\[ n = 24 \]

So, there are 24 terms in the series.

Now, we can express \( P \) as:
\[ P = \sum_{k=1}^{24} (2 + (k-1) \cdot 4)^2 \]

Calculating each term:
1. The \( k \)-th term is \( (4k - 2)^2 \).
2. Expanding this:
\[ P = \sum_{k=1}^{24} (4k - 2)^2 = \sum_{k=1}^{24} (16k^2 - 16k + 4) \]
\[ P = 16 \sum_{k=1}^{24} k^2 - 16 \sum_{k=1}^{24} k + 4 \cdot 24 \]

Using the formulas for the sums:
- \( \sum_{k=1}^{n} k = \frac{n(n+1)}{2} \)
- \( \sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6} \)

For \( n = 24 \):
- \( \sum_{k=1}^{24} k = \frac{24 \cdot 25}{2} = 300 \)
- \( \sum_{k=1}^{24} k^2 = \frac{24 \cdot 25 \cdot 49}{6} = 4900 \)

Now substituting back into the equation for \( P \):
\[ P = 16 \cdot 4900 - 16 \cdot 300 + 96 \]
\[ P = 78400 - 4800 + 96 = 73696 \]

### Step 2: Calculate \( Q \)

The series for \( Q \) is given by:
\[ Q = 1^2 + 5^2 + 9^2 + \ldots + 81^2 \]

This series consists of squares of numbers that form another arithmetic sequence where the first term \( a = 1 \) and the common difference \( d = 4 \).

To find the number of terms in this series, we set \( a_n = 81 \):
\[ 81 = 1 + (n-1) \cdot 4 \]
\[ 80 = (n-1) \cdot 4 \]
\[ n-1 = 20 \]
\[ n = 21 \]

So, there are 21 terms in the series.

Now we can express \( Q \) as:
\[ Q = \sum_{k=1}^{21} (1 + (k-1) \cdot 4)^2 \]

Calculating each term:
1. The \( k \)-th term is \( (4k - 3)^2 \).
2. Expanding this:
\[ Q = \sum_{k=1}^{21} (4k - 3)^2 = \sum_{k=1}^{21} (16k^2 - 24k + 9) \]
\[ Q = 16 \sum_{k=1}^{21} k^2 - 24 \sum_{k=1}^{21} k + 21 \cdot 9 \]

For \( n = 21 \):
- \( \sum_{k=1}^{21} k = \frac{21 \cdot 22}{2} = 231 \)
- \( \sum_{k=1}^{21} k^2 = \frac{21 \cdot 22 \cdot 43}{6} = 3311 \)

Now substituting back into the equation for \( Q \):
\[ Q = 16 \cdot 3311 - 24 \cdot 231 + 189 \]
\[ Q = 52976 - 5544 + 189 = 47421 \]

### Step 3: Calculate \( P - Q \)

Now we can find \( P - Q \):
\[ P - Q = 73696 - 47421 = 26275 \]

### Final Answer
The value of \( P - Q \) is \( 26275 \).