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प्रश्न
The sum of squares of first n natural numbers is given by `1/6n(n + 1)(2n + 1)` or `1/6(2n^3 + 3n^2 + n)`. Find the sum of squares of the first 10 natural numbers.
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उत्तर
Given, the sum of squares of first n natural numbers = `1/6(n +1)(2n + 1)`
∴ The sum of squares of first 10 natural numbers ...[∴ Put n = 10]
= `1/6 (10)(10 + 1)(2 xx 10+1)`
= `1/6 xx 10 xx 11 xx 21`
= 385
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संबंधित प्रश्न
Observe the patterns of digits made from line segments of equal length. You will find such segmented digits on the display of electronic watches or calculators.
If the number of digits formed is taken to be n, the number of segments required to
form n digits is given by the algebraic expression appearing on the right of each pattern.
How many segments are required to form 5, 10, 100 digits of the kind −
Use the given algebraic expression to complete the table of number patterns.
| S. No |
Expression |
Terms | |||||||||
| 1st | 2nd | 3rd | 4th | 5th | ... | 10th | ... | 100th | ... | ||
| 1 | 2n - 1 | 1 | 3 | 5 | 7 | 9 | - | 19 | - | - | - |
| 2 | 3n + 2 | 5 | 8 | 11 | 14 | - | - | - | - | - | - |
| 3 | 4n + 1 | 5 | 9 | 13 | 17 | - | - | - | - | - | - |
| 4 | 7n + 20 | 27 | 34 | 41 | 48 | - | - | - | - | - | - |
| 5 | n2 + 1 | 2 | 5 | 10 | 17 | - | - | - | - | 10001 | - |
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