Find the sum of n terms of 1+(1+2)/2+(1+2+3)/3+⋯
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EXPLANATION.
Sum of the n terms,
⇒ 1 + (1 + 2)/2 + (1 + 2 + 3)/3 +,,,,,,,,,,,n terms.
As we know that,
Nth terms = 1 + 2 + 3 + 4 + 5,,,,,,n/n.
Sum of first n natural numbers = ∑r = n(n + 1)/2.
⇒ n(n + 1)/2/n.
⇒ (n + 1)/2.
⇒ ∑(n + 1)/2.
⇒ ∑n/2 + 1/2.
⇒ 1/2∑n + ∑1/2.
⇒ 1/2 X n(n + 1)/2 + n/2.
⇒ n² + n/4 + n/2.
⇒ n² + n + 2n/4.
⇒ n² + 3n/4.
MORE INFORMATION.
Supposition of terms in A.P.
(1) = Three terms as = a - d, z, a + d.
(2) = Five terms as = a - 2d, a - d, a, a + d, a + 2d.
(3) = Four terms as = a - 3d, a - d, a + d, a + 3d.
Supposition of terms in G.P.
(1) = Three terms as = a/r, a, ar.
(2) = Five terms as = a/r², a/r, a, ar, ar².
(3) = Four terms as = a/r³, a/r, ar, ar³.
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