If we combine two consecutive triangular numbers, we get?

Let the two consecutive triangular numbers be T_n = n(n+1)/2 and T_{n+1} = (n+1)(n+2)/2.

Their sum is:

T_n + T_{n+1} = n(n+1)/2 + (n+1)(n+2)/2

Factor out (n+1)/2:

(n+1)/2 * (n + n+2) = (n+1)/2 * (2n+2)

Simplify:

(n+1)/2 * 2(n+1) = (n+1)²

Thus, the sum is (n+1)², which is a perfect square.

हिन्दी उत्तर:

मान लीजिए दो क्रमागत त्रिभुज संख्याएँ T_n = n(n+1)/2 और T_{n+1} = (n+1)(n+2)/2 हैं।

उनका योग है:

T_n + T_{n+1} = n(n+1)/2 + (n+1)(n+2)/2

(n+1)/2 को बाहर निकालें:

(n+1)/2 * (n + n+2) = (n+1)/2 * (2n+2)

सरल करें:

(n+1)/2 * 2(n+1) = (n+1)²

इसलिए, योग (n+1)² है, जो एक पूर्ण वर्ग है।