Question

find the sum to n terms of the series 5+14+19+29+41+.....​

Answer 1

5+14+19+29+41+.....​ =   n(n+2)(n+4)/3

Step-by-step explanation:

Correct Question would be

5 + 11  + 19  + 29  + 41 +.........................

= (2² + 1  )  + (3² + 2)  + (4² + 3)  + (5² + 4) + (6²+ 5) +.................((n + 1)² + n)

= (2² + 3²  + 4² + 5² +..................................+ (n + 1)²)  + (1 + 2 + 3 +.........+ n)

= (1² + 2² + 3²  + 4² + 5² +..................................+ (n + 1)²) ) - 1  + n(n+1)/2

∑r² = r(r + 1)(2r + 1)/6

r = n + 1

= (n + 1)(n + 2)(2n + 3)/6  - 1  + n(n+1)/2

= ((n + 1)(n + 2)(2n + 3)  - 6   + 3n(n+1) )/6

= ((n² + 3n + 2)(2n + 3)  - 6  + 3n(n+1) )/6

= (2n³ + 9n² + 13n  + 6 - 6 + 3n²  + 3n)/6

= (2n³  + 12n² + 16n)/6

= 2n(n² + 6n  + 8)/6

= n(n² + 6n  + 8)/3

=  n(n² + 4n + 2n  + 8)/3

= n(n+2)(n+4)/3

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find the sum to n terms of the series

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