Find the value of 1 + 3 + 5 + … + 27, that is, all the odd numbers from 1 up to 27.
Class 8Squares and Square RootsSquares and Roots
A169
B729
C378
D196
Step-by-Step Solution
The last term 27=2n−1 gives n=14, so there are 14 odd numbers and the sum is 142=196. Choosing 169 counts only 13 terms; 729 is 272, squaring the last term instead of counting the terms; and 378 is 1 + 2 + 3 + … + 27, which wrongly includes the even numbers.
Answer: D) 196
Key idea: Square root by long division
Use this for larger numbers (or non-perfect squares).
**Step 1**: Pair the digits from the right (e.g. 1849 → 18 49). **Step 2**: Find the largest integer whose square ≤ first pair. Write it as the first quotient digit. Subtract its square. **Step 3**: Bring down the next pair. Double the current quotient — that's your divisor's prefix. Find a digit d such that (prefix × 10+d)×d ≤ current dividend. **Step 4**: Append d to quotient, subtract, repeat.
Example: 1849=43.(42=16 ≤ 18; remainder 2; bring down 49 → 249; double 4 = 8; find d where (80+d)d ≤ 249; d=3 gives 83×3=249.) Quotient = 43.
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