JEE Mathematics Question

Given: r=1^15 r^2 15r = 2^m 3^n 5^k Find: m+n+k. Use the identity r^2 15r = 15r 14r-1 and then split r = (r-1)+1. Equivalently, r^2 15r = 15(r-1) 14r-1 + 15 14r-1 which gives r^2 15r = 15 14 13r-2 + 15 14r-1. Therefore, r=1^15 r^2 15r = 15 14 r=1^15 13r-2 + 15 r=1^15 14r-1. Now use the binomial sums r=1^15 13r-2 = 2^13, r=1^15 14r-1 = 2^14. So, r=1^15 r^2 15r = 15 14 2^13 + 15 2^14. Factoring, = 15 2^13(14+2) = 15 2^13 16 = 15 2^17. Now, 15 2^17 = 3^1 5^1 2^17. Hence, m=17, n=1, k=1. Therefore, m+n+k = 17+1+1 = 19. The correct option is A. From the solution, the series is interpreted as 1^2 151+2^2 152+3^2 153+ +15^2 1515. Write r^2 = r(r-1)+r. Then r=1^15 r^2 15r = r=1^15 r(r-1) 15r + r=1^15 r 15r. Now apply the identities r(r-1) 15r = 15 14 13r-2, and r 15r = 15 14r-1. So, r=1^15 r^2 15r = 15 14 r=1^15 13r-2 + 15 r=1^15 14r-1. Each sum is a complete binomial sum after index shift: r=1^15 13r-2 = t=0^13 13t = 2^13, r=1^15 14r-1 = u=0^14 14u = 2^14. Hence, r=1^15 r^2 15r = 15 14 2^13 + 15 2^14 = 15 2^17. Prime factorizing, 15 2^17 = 2^173^15^1. Thus m=17, n=1, k=1, so m+n+k=19.

JEE Question — Solution | JEE Mathematics

MCQMediumJEE 2025Simple Applications

Expression asking: If 1 squared times 15C1 plus 2 squared times 15C2 plus 3 squared times 15C3 up to 15 squared times 15C15 equals 2 raised to m times 3 raised to n times 5 raised to k, find m plus n plus k.

A
$19$
B
$21$
C
$18$
D
$20$

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given:

\sum_{r=1}^{15} r^2 \binom{15}{r} = 2^m 3^n 5^k

Find: $m+n+k$ .

Use the identity

r^2 \binom{15}{r} = 15r \binom{14}{r-1}

and then split $r = (r-1)+1$ . Equivalently,

r^2 \binom{15}{r} = 15(r-1)\binom{14}{r-1} + 15\binom{14}{r-1}

which gives

r^2 \binom{15}{r} = 15\cdot 14 \binom{13}{r-2} + 15\binom{14}{r-1}.

Therefore,

\sum_{r=1}^{15} r^2 \binom{15}{r} = 15\cdot 14 \sum_{r=1}^{15} \binom{13}{r-2} + 15 \sum_{r=1}^{15} \binom{14}{r-1}.

Now use the binomial sums

\sum_{r=1}^{15} \binom{13}{r-2} = 2^{13}, \qquad \sum_{r=1}^{15} \binom{14}{r-1} = 2^{14}.

So,

\sum_{r=1}^{15} r^2 \binom{15}{r} = 15\cdot 14\cdot 2^{13} + 15\cdot 2^{14}.

Factoring,

= 15\cdot 2^{13}(14+2) = 15\cdot 2^{13}\cdot 16 = 15\cdot 2^{17}.

Now,

15\cdot 2^{17} = 3^1 \cdot 5^1 \cdot 2^{17}.

Hence,

m=17, \; n=1, \; k=1.

Therefore, $m+n+k = 17+1+1 = 19$ . The correct option is A.

Using standard binomial identities

From the solution, the series is interpreted as

1^2\binom{15}{1}+2^2\binom{15}{2}+3^2\binom{15}{3}+\cdots+15^2\binom{15}{15}.

Write

r^2 = r(r-1)+r.

Then

\sum_{r=1}^{15} r^2 \binom{15}{r} = \sum_{r=1}^{15} r(r-1)\binom{15}{r} + \sum_{r=1}^{15} r\binom{15}{r}.

Now apply the identities

r(r-1)\binom{15}{r} = 15\cdot 14\binom{13}{r-2},

and

r\binom{15}{r} = 15\binom{14}{r-1}.

So,

\sum_{r=1}^{15} r^2 \binom{15}{r} = 15\cdot 14 \sum_{r=1}^{15}\binom{13}{r-2} + 15\sum_{r=1}^{15}\binom{14}{r-1}.

Each sum is a complete binomial sum after index shift:

\sum_{r=1}^{15}\binom{13}{r-2} = \sum_{t=0}^{13}\binom{13}{t} = 2^{13},

\sum_{r=1}^{15}\binom{14}{r-1} = \sum_{u=0}^{14}\binom{14}{u} = 2^{14}.

Hence,

\sum_{r=1}^{15} r^2 \binom{15}{r} = 15\cdot 14\cdot 2^{13} + 15\cdot 2^{14} = 15\cdot 2^{17}.

Prime factorizing,

15\cdot 2^{17} = 2^{17}3^15^1.

Thus $m=17$ , $n=1$ , $k=1$ , so $m+n+k=19$ .

Common mistakes

Using $\sum r^2\binom{n}{r}$ directly without first rewriting $r^2$ as $r(r-1)+r$ . This is wrong because the standard binomial identities apply to $r\binom{n}{r}$ and $r(r-1)\binom{n}{r}$ separately. Rewrite the term before summing.
Missing the second term after decomposition and taking only $15\cdot 14\cdot 2^{13}$ . This is wrong because $r^2 = r(r-1)+r$ has two contributions. Add both sums before factorization.
Making an index-shift error in $\sum \binom{13}{r-2}$ or $\sum \binom{14}{r-1}$ . This is wrong because the shifted limits produce complete binomial sums. Convert carefully to $\sum_{t=0}^{13}\binom{13}{t}$ and $\sum_{u=0}^{14}\binom{14}{u}$ .

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